Diamond tetrahedral

Rgure 4.58 Structure of diamond tetrahedral unit and lattice... 

This is a relatively rare structure, diamond being probably the best known example. Here, the carbon atoms are not close-packed. Each carbon is surrounded tetrahedrally by four other carbon atoms (Figure 2.1). Clearly, each carbon is exerting a tetrahedrally directed... 

Crystalline silicon has the tetrahedral diamond arrangement, but since the mean thermochemical bond strength between the silicon atoms is less than that found between carbon atoms (Si—Si, 226 kJmol , C—C, 356kJmol ), silicon does not possess the great hardness found in diamond. Amorphous silicon (silicon powder) is microcrystalline silicon. 

Fig. 8.7 Cubic and tetrahedral (diamond) lattices, which are commonly used for lattice simulations of polymers.

Silicon atoms bond strongly with four oxygen atoms to give a tetrahedral unit (Fig. 16.4a). This stable tetrahedron is the basic unit in all silicates, including that of pure silica (Fig. 16.3c) note that it is just the diamond cubic structure with every C atom replaced by an Si04 unit. But there are a number of other, quite different, ways in which the tetrahedra can be linked together. 

Diamond is an important commodity as a gemstone and as an industrial material and there are several excellent monographs on the science and technology of this material [3-5]. Diamond is most frequently found in a cubic form in which each carbon atom is linked to fom other carbon atoms by sp ct bonds in a strain-free tetrahedral array. Fig. 2A. The crystal stmcture is zinc blende type and the C-C bond length is 154 pm. Diamond also exists in an hexagonal form (Lonsdaleite) with a wurtzite crystal structure and a C-C bond length of 152 pm. The crystal density of both types of diamond is 3.52 g-cm. ... 

Figure 8.3 Structure of diamond showing the tetrahedral coordination of C the dashed lines indicate the cubic unit cell containing 8 C atoms.

Figure 29.1 Crystal structures of ZnS. (a) Zinc blende, consisting of two, interpenetrating, cep lattices of Zn and S atoms displaced with respect to each other so that the atoms of each achieve 4-coordination (Zn-S = 235 pm) by occupying tetrahedral sites of the other lattice. The face-centred cube, characteristic of the cep lattice, can be seen — in this case composed of S atoms, but an extended diagram would reveal the same arrangement of Zn atoms. Note that if all the atoms of this structure were C, the structure would be that of diamond (p. 275). (b) Wurtzite. As with zinc blende, tetrahedral coordination of both Zn and S is achieved (Zn-S = 236 pm) but this time the interpenetrating lattices are hexagonal, rather than cubic, close-packed.

In diamond, each carbon atom forms single bonds with four other carbon atoms arranged tetrahedrally around it The hybridization in diamond is sp3. The three-dimensional covalent bonding contributes to diamond s unusual hardness. Diamond is one of the hardest substances known it is used in cutting tools and quality grindstones (Figure 9.12). 

The structure of a diamond. Diamond has a three-dimensional structure in which each carbon atom is surrounded tetrahedrally by four other carbon atoms. 

Furthermore, spz bonding is connected with tetrahedral bond angles (as in Figure 16-11). These expectations are consistent with the experimentally determined structure of diamond, shown in Figure 17-2. 

FIGURE 14.30 Structure of diamond. Each carbon atom is sp hybridized and forms tetrahedral rr-bonds to four neighbors. This pattern is repeated throughout the crystal and accounts for diamond s great hardness. 

In diamond, each carbon atom is sp3 hybridized and linked tetrahedrally to its four neighbors, with all electrons in C C cr-bonds (Fig. 14.30). Diamond is a rigid, transparent, electrically insulating solid. It is the hardest substance known and the best conductor ol heat, being about five times better than copper. These last two properties make it an ideal abrasive, because it can scratch all other substances, yet the heat generated by friction is quickly conducted away. 

In diamond, carbon is sp hybridized and forms a tetrahedral, three-dimensional network structure, which is extremely rigid. Graphite carbon is sp2 hybridized and planar. Its application as a lubricant results from the fact that the two-dimensional sheets can slide across one another, thereby reducing friction. In graphite, the unhybridized p-electrons are free to move from one carbon atom to another, which results in its high electrical conductivity. In diamond, all electrons are localized in sp3 hybridized C—C cr-bonds, so diamond is a poor conductor of electricity. 

The Tetrahedral Carbon Atom.—We have thus derived the result that an atom in which only s and p eigenfunctions contribute to bond formation and in which the quantization in polar coordinates is broken can form one, two, three, or four equivalent bonds, which are directed toward the corners of a regular tetrahedron (Fig. 4). This calculation provides the quantum mechanical justification of the chemist s tetrahedral carbon atom, present in diamond and all aliphatic carbon compounds, and for the tetrahedral quadrivalent nitrogen atom, the tetrahedral phosphorus atom, as in phosphonium compounds, the tetrahedral boron atom in B2H6 (involving single-electron bonds), and many other such atoms. 

Thus we have shown that when s and p orbitals are available and s—p quantization is broken an atom can form four (or fewer) equivalent bonds which are directed towards tetrahedron corners. To the approximation involved in these calculations the strength of a bond is independent of the nature of other bonds. This result gives us at once the justification for the tetrahedral carbon atom and other tetrahedral atoms, such as silicon, germanium, and tin in the diamond-type crystals of the elements and, in general, all atoms in tetrahedral structures. 

In this discussion of the transition elements we have considered only the orbitals (n— )d ns np. It seems probable that in some metals use is made also of the nd orbitals in bond formation. In gray tin, with the diamond structure, the four orbitals 5s5p3 are used with four outer electrons in the formation of tetrahedral bonds, the 4d shell being filled with ten electrons. The structure of white tin, in which each atom has six nearest neighbors (four at 3.016A and two at 3.17.5A), becomes reasonable if it is assumed that one of the 4d electrons is promoted to the 5d shell, and that six bonds are formed with use of the orbitals 4dSs5p35d. 

The arrangement of the centers of the molecules in the crystal is that corresponding to the diamond structure. Each molecule is surrounded tetrahedrally by four molecules. If we consider a molecule as roughly tetrahedral in shape with similar orientation to the tetrahedron formed by the four beryllium atoms, then the adjacent molecules are so oriented as to present tetrahedral faces to one another. 

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