Pearson symbol: cF8, prototype: C. Four elements, from column IV
of the periodic table, crystallize in the diamond structure, namely: carbon, silicon, germanium and gray tin (which is one of the two allotropes of tin at normal pressure and temperature). The atoms of each of these elements have four electrons in the outermost shell (the so called valence shell). By completing this shell with four additional electrons those atoms can achieve a state of the highest stability. This stability is reached in the crystal of each of these elements in which an atom is surrounded by four neighboring atoms that in turn form covalent chemical bonds (represented schematically in Fig. 90) with it. In the diamond structure, each atom shares four electrons with its 4 NNs and each of these neighbors shares an electron with the atom under consideration. Therefore, all atoms can complete the 4 electrons that were lacking to achieve the highest stability.
The neighborhood of an atom in the diamond structure is shown in Fig. 91. The four
structure are placed
under consideration in the center, like
Figure 90 Two-dimensional schematic representation of covalent chemical bonds in the
diamond structure.
Figure 91 (a) A tetrahedron defined by the
tetrahedron from (a) inscribed in a cube. covalent bonds between an atom and its 4
The neighborhood of an atom in the diamond structure is shown in Fig. 91. The four NNs of each atom of an element that crystallize in this structure are placed at the vertices of a regular tetrahedron that has the atom under consideration in the center, like it is shown in Fig. 91a. The regular dimensional schematic representation of covalent chemical bonds in the
A tetrahedron defined by the NNs of an atom in the diamond structure. tetrahedron from (a) inscribed in a cube. (c) Three-dimensional schematic representation of covalent bonds between an atom and its 4 NNs.
The neighborhood of an atom in the diamond structure is shown in s of each atom of an element that crystallize in this
s of a regular tetrahedron that has the atom is shown in Fig. 91a. The regular dimensional schematic representation of covalent chemical bonds in the
s of an atom in the diamond structure. (b) The ematic representation of
tetrahedron is easier to draw if we place it inside a cube, what was done in Fig. 91b. In addition, Fig. 91c shows a three-dimensional schematic representation of covalent bonds between an atom in the diamond structure and its 4 NNs.
We should observe that the cubic volume that we have drawn in Fig. 91b does not, of course, represent a unit cell of the diamond structure,
since it does not have atoms in all its vertices. However, we can easily
locate it within the cubic unit cell of this structure. Figure 92 shows two of the 4 possible positions of the small cube inside the diamond cubic unit cell. We may also observe in Fig. 92 that this unit cell is just the cubic unit
cell of the fcc structure with 4 additional atoms placed inside (on the body diagonals). The distance between each additional atom and its nearest cube vertex is 1 4 of the cube body diagonal, and those additional atoms occupy tetrahedral interstices present in the fcc cubic unit cell. We can see in Fig. 92 that in this cell, there is a total of 8 tetrahedral interstices and in the case of the diamond structure half of them are filled with atoms.
The tetrahedral interstices present in the fcc structure have been already considered by us in the previous section. In that opportunity, the fcc structure was seen as a sequence of two-dimensional hcp layers of
ABCABC… type. In Fig. 89a, we have shown two examples of tetrahedral
interstices present in the cubic cell of the fcc structure.
Figure 92 Two small cubes from Fig. 91b placed in two of the four possible positions inside
To help visualize the position
unit cell of the diamond structure, we have dra orthogonal vertical planes
diagonals of the cube and the 4 atoms are placed on these diagonals in the way explained in this figure.
It is obvious that the neighborhood of each atom in the diamond structure is the same. This can be verified by drawing two cubic unit cells, I and II, in the diamond structure in such a way that cube II
respect to cube I along one of its body diagonals, to a segment equal in length to 1 4 of the diagonal length. We obtain then that the atoms that are in the interior of cube I coincide with the vertices or centers of
cube II, thus suc
atoms on the diagonals of cube I. This is illustrated in 2D in Fig. 94, where we have plotted a plane with 12 atoms from certain region of the crystal. This plane includes
diagonals. One of these atoms is placed at a vertex of cube II and the other one at its face center. We can see in Fig. 94a that the atoms that are located on the body diagonals of the cubes have the same s
atoms from vertices and faces of cube I. It is also easy to observe in Fig. 94b the equivalence between the relative distributions of atoms of each
Figure 93 Relative positions
are inside the cubic unit cell are distributed in two vertic diagonals of the cube.
To help visualize the positions of the 4 atoms that are inside of the cubic e diamond structure, we have drawn in Fig. 93 two mutually orthogonal vertical planes A and B. Each plane is defined by two body cube and the 4 atoms are placed on these diagonals in the way explained in this figure.
It is obvious that the neighborhood of each atom in the diamond structure is the same. This can be verified by drawing two cubic unit cells, I and II, in the diamond structure in such a way that cube II is shifted with respect to cube I along one of its body diagonals, to a segment equal in of the diagonal length. We obtain then that the atoms that are in the interior of cube I coincide with the vertices or centers of the faces of ch atoms of cube II have the same neighborhood as the atoms on the diagonals of cube I. This is illustrated in 2D in Fig. 94, where we have plotted a plane with 12 atoms from certain region of the crystal. This plane includes a cross section of cube I with two atoms from the diagonals. One of these atoms is placed at a vertex of cube II and the other one at its face center. We can see in Fig. 94a that the atoms that are located on the body diagonals of the cubes have the same spatial distribution as the atoms from vertices and faces of cube I. It is also easy to observe in Fig. 94b the equivalence between the relative distributions of atoms of each
Relative positions of atoms belonging to the diamond structure. The 4 atoms that are inside the cubic unit cell are distributed in two vertical planes defined by the body diagonals of the cube.
4 atoms that are inside of the cubic two mutually . Each plane is defined by two body cube and the 4 atoms are placed on these diagonals in the It is obvious that the neighborhood of each atom in the diamond structure is the same. This can be verified by drawing two cubic unit cells, I is shifted with respect to cube I along one of its body diagonals, to a segment equal in of the diagonal length. We obtain then that the atoms that are the faces of have the same neighborhood as the atoms on the diagonals of cube I. This is illustrated in 2D in Fig. 94, where we have plotted a plane with 12 atoms from certain region of the crystal. a cross section of cube I with two atoms from the diagonals. One of these atoms is placed at a vertex of cube II and the other one at its face center. We can see in Fig. 94a that the atoms that are located patial distribution as the
atoms from vertices and faces of cube I. It is also easy to observe in Fig. 94b the equivalence between the relative distributions of atoms of each
of atoms belonging to the diamond structure. The 4 atoms that al planes defined by the body
type (from the body diagonals and from the vertices and faces).
can say that it is the atomic arrangement in the diamond structure which
Figure 94 (a) Comparison of the distribution of atoms from v
unit cells with those from their body diagonals in the diamond structure. atoms (considered hard spheres) from (a)
the relative distributions of atoms o those from its diagonals) is visualized.
(from the body diagonals and from the vertices and faces).Finally, we can say that it is the atomic arrangement in the diamond structure which Comparison of the distribution of atoms from vertices and faces of the cubic unit cells with those from their body diagonals in the diamond structure. (b) Cross sections of atoms (considered hard spheres) from (a) are shown. In this figure, the equivalency between the relative distributions of atoms of each type (those from vertices and faces of the cube and those from its diagonals) is visualized.
Finally, we can say that it is the atomic arrangement in the diamond structure which ertices and faces of the cubic Cross sections of shown. In this figure, the equivalency between f each type (those from vertices and faces of the cube and
allows each atom to be in the middle of a regular tetrahedron with 4 (located at the vertices of the tetrahedron) that are covalently bonded to it.
To conclude, we can say that the diamond structure is just a superposition of two
the other in the way described above. Each substructure may be seen as a sequence of layers of
shifted one with respect to the other, to a segment shorter than the distance
Figure 95 Left part of Fig. 94b with the
Figure 96 Cubic and rhombohedral unit cells for the diamond structure (left).
is also shown the position
each atom to be in the middle of a regular tetrahedron with 4 (located at the vertices of the tetrahedron) that are covalently bonded to it.
To conclude, we can say that the diamond structure is just a superposition of two fcc substructures that are shifted one with respect to the other in the way described above. Each substructure may be seen as a sequence of layers of ABCABC… type and the two substructures are shifted one with respect to the other, to a segment shorter than the distance
Left part of Fig. 94b with the cross sections of the A, B, and C layer planes adde
Cubic and rhombohedral unit cells for the diamond structure (left). In the figure, it he positions of the two atoms belonging to the rhombohedral unit cell (right).
each atom to be in the middle of a regular tetrahedron with 4 NNs (located at the vertices of the tetrahedron) that are covalently bonded to it.
To conclude, we can say that the diamond structure is just a substructures that are shifted one with respect to the other in the way described above. Each substructure may be seen as a substructures are shifted one with respect to the other, to a segment shorter than the distance planes added.
In the figure, it of the two atoms belonging to the rhombohedral unit cell (right).
between two consecutive layers in the substructures, in the direction orthogonal to the layer planes. Thus, the diamond structure represents indeed a sequence
see in Fig. 95, where we have added the cross sections of layer planes to Fig. 94b.
The smallest unit cell of the diamond structure the primitive rhombohedral unit cel
cell contains 2 atoms as shown in Fig. 96. Therefore, the diamond structure can be seen as a fcc
Finally, let us show the position cubic unit cell. Fig
cell and Fig. 97b shows the projection of these atoms on the cell base. The coordinates of the 8 atoms, in Fig. 97a, are given in terms of the cub
a, b, c, and the fractions near the atom projections in Fig. 97b represent the coordinates of these atoms in the