91Ó°ÊÓ

The alkali metal fulleride superconductors \(\mathrm{M}_{3} \mathrm{C}_{60}\) have a cubic closest-packed (face-centered cubic) arrangement of nearly spherical \(\mathrm{C}_{60}{ }^{3-}\) anions with \(\mathrm{M}^{+}\) cations in the holes between the larger \(\mathrm{C}_{60}{ }^{3-}\) ions. The holes are of two types: octahedral holes, which are surrounded octahedrally by six \(\mathrm{C}_{60}{ }^{3-}\) ions, and tetrahedral holes, which are surrounded tetrahedrally by four \(\mathrm{C}_{60}{ }^{3-}\) ions. (a) Sketch the three-dimensional structure of one unit cell. (b) How many \(\mathrm{C}_{60}{ }^{3-}\) ions, octahedral holes, and tetrahedral holes are present per unit cell? (c) Specify fractional coordinates for all the octahedral and tetrahedral holes. (Fractional coordinates are fractions of the unit cell edge lengths. For example, a hole at the center of the cell has fractional coordinates \(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\).) (d) The radius of a \(\mathrm{C}_{60}{ }^{3-}\) ion is about \(500 \mathrm{pm}\). Assuming that the \(\mathrm{C}_{60}{ }^{3-}\) ions are in contact along the face diagonals of the unit cell, calculate the radii of the octahedral and tetrahedral holes. (e) The ionic radii of \(\mathrm{Na}^{+}, \mathrm{K}^{+}\), and \(\mathrm{Rb}^{+}\) are 102,138 , and \(152 \mathrm{pm}\), respectively. Which of these ions will fit into the octahedral and tetrahedral holes? Which ions will fit only if the framework of \(\mathrm{C}_{60}{ }^{3-}\) ions expands?

Step by step solution

01

Understand the Unit Cell

The face-centered cubic (fcc) structure contains atoms at the corners and the centers of each face of the cube. Each corner atom is shared by eight adjacent unit cells, and each face atom is shared by two.

02

Count the Number of Cations and Anions

In an FCC structure, there are 4 atoms per unit cell. The diagram includes 8 corners and 6 face-centered positions. Each contributes as follows: corners = (1/8)x8 = 1, and face-centered = (1/2)x6 = 3, totaling 4 C60^{3-} ions. Octahedral and tetrahedral holes match the number of anions.

03

Identify Octahedral and Tetrahedral Holes

Octahedral holes are located at the centers of edges and the unit cell center, totaling 4 per cell due to symmetry and hole sharing (12 edge holes and 1 center hole). Tetrahedral holes are fully contained within each cell and situated near the quarter divisions of cell edges, totaling 8 per unit cell.

04

Determine Fractional Coordinates

Octahedral holes are at (0.5, 0.5, 0.5) and centers of edges like (0.5, 0, 0); tetrahedral holes are at (0.25, 0.25, 0.25), (0.75, 0.75, 0.75), etc.

05

Calculate Radii of Holes

The face-diagonal length in terms of radius is \(4r = a \sqrt{2}\). Setting \(r = 500\, \text{pm}\), solve for \(a\) (unit cell edge length). Octahedral hole radius is \(a\sqrt{3}/2 - 500\, \text{pm}\), and tetrahedral hole radius is \(a \sqrt{2}/4 - 500 \text{pm}\).

06

Determine Suitable Cations

Calculate hole radii and compare to cation radii. Only ions like Na^+ may fit in tetrahedral holes without expansion; K^+ and Rb^+ may fit octahedrally if C60^{3-} allows for expansion.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Unit Cell

In crystallography, understanding the concept of a unit cell is essential. The unit cell is the fundamental building block of a crystal's structure. It is the smallest, repeating portion of the crystal lattice that shows the entire symmetry of the crystal structure.
In our example, the alkali metal fulleride superconductors involve a face-centered cubic (fcc) arrangement. In this setup, the unit cell is cubic in shape, with atoms located at each of its corners and at the centers of each of the cube’s faces. A traditional unit cell includes:

• 8 atoms at the corners, where each corner atom is shared by eight adjacent unit cells which count as a total of 1 full atom per unit cell.
• 6 atoms at the face centers, where each face atom is shared by two unit cells which contribute as 3 full atoms per unit cell.

These components result in a total of 4 atoms per face-centered cubic unit cell.

Face-Centered Cubic

The face-centered cubic (fcc) structure is one of the most common crystal arrangements. It is density-efficient, allowing atoms to pack closely together. In a face-centered cubic crystal:

• Atoms are at each of the eight corners of the cube.
• There is also an atom in the center of each of the cube's six faces.

Each corner atom is shared by eight neighboring unit cells, while each face-centered atom is shared by two. This atom sharing results in four atoms per fcc unit cell. This efficient packing creates a highly symmetrical structure, facilitating the presence of holes where smaller ions can fit. In the context of alkali metal fulleride superconductors, the \(C_{60}^{3-}\) anions form this face-centered cubic lattice.

Octahedral Holes

Inside the face-centered cubic structure are octahedral holes, which are vacant spaces formed between the larger atoms of the \(C_{60}^{3-}\) lattice. These holes occur at:

• The center of the unit cell.
• The centers of the edges of the unit cell, altogether providing a total of four octahedral holes due to symmetry and adjacent sharing.

Imagine each octahedral hole being surrounded by six \(C_{60}^{3-}\) ions, similar to the shape of an octahedron. These sites are crucial because they can hold cations, playing an important role in the properties of materials, such as conductivity and stability.

Tetrahedral Holes

Tetrahedral holes are another significant feature in the face-centered cubic unit cell. These smaller spaces are located at positions close to corners, lying fully within the individual unit cells. Each unit cell contains:

• Four tetrahedral holes making eight complete tetrahedral sites within the cell due to full ownership by the cell.

These holes are smaller compared to octahedral holes, formed by being surrounded by four \(C_{60}^{3-}\) ions in a tetrahedral geometry, like a pyramid. Due to their location and size, choosing smaller cations that can fit into these holes is necessary, and only certain ions will fit without causing expansion of the surrounding lattice. This positioning and dimension of holes is a determining factor when analyzing which ions can occupy these spaces without causing structural changes.