Metal simple cubic structure

It has been suggested that a brown form of P, formed by condensing the vapour of white P at 1000°C or the vapour of red P at 350°C on to a cold finger at — 196°C, may consist of Pj molecules. A vitreous form of P has been prepared by heating white P with Hg at 380°C at 450°C black P is formed. The conversion of P under pressure to forms with the As structure (at 80 kbar) and a (metallic) simple cubic structure (at 110 kbar) has been reported, but details are not available. 

Although the comer atoms must move apart to convert a simple cube into a body-centered cube, the extra atom in the center of the stracture makes the body-centered cubic lattice more compact than the simple cubic structure. All the alkali metals, as well as iron and the transition metals from Groups 5 and 6, form ciystals with body-centered cubic structures. 

We can determine the amount of empty space in the simple cubic (a space-filling model is shown in Figure 7.15) structure by considering it to have an edge length l, which will be twice the radius of an atom. Therefore, the radius of the atom is 1/2, so the volume of one atom is (4/3)7r(l/2)3 = 0.52413, but the volume of the cube is P. From this we see that because the cube contains only one atom that occupies 52.4% of the volume of the cube, there is 47.6% empty space. Because of the low coordination number and the large amount of empty space, the simple cubic structure does not represent an efficient use of space and does not maximize the number of metal atoms bonded to each other. Consequently, the simple cubic structure is not a common one for metals. 

According to Pearson (1972) the rhombohedral structure of these elements can be considered a distortion of a simple cubic structure in which the d2/d ratio would be 1. The decrease of the ratio on passing from As to Bi, and the corresponding relative increase of the strength of the X-X interlayer bond (passing from a coordination nearly 3, as for the 8 — eat rule, to a coordination closer to 6) can be related to an increasing metallic character. 

We now need to define a collection of atoms that can be used in a DFT calculation to represent a simple cubic material. Said more precisely, we need to specify a set of atoms so that when this set is repeated in every direction, it creates the full three-dimensional crystal stmcture. Although it is not really necessary for our initial example, it is useful to split this task into two parts. First, we define a volume that fills space when repeated in all directions. For the simple cubic metal, the obvious choice for this volume is a cube of side length a with a corner at (0,0,0) and edges pointing along the x, y, and z coordinates in three-dimensional space. Second, we define the position(s) of the atom(s) that are included in this volume. With the cubic volume we just chose, the volume will contain just one atom and we could locate it at (0,0,0). Together, these two choices have completely defined the crystal structure of an element with the simple cubic structure. The vectors that define the cell volume and the atom positions within the cell are collectively referred to as the supercell, and the definition of a supercell is the most basic input into a DFT calculation. 

Assume there exists some hypothetical metal that exhibits ferromagnetic behavior and that has a simple cubic structure, an atomic radius of 0.153 nm, and a saturation flux density of 0.76 tesla. Determine the number of Bohr magnetons per atom for this material. 

All P, O, and T layers have the same hexagonal close-packed arrangement within each layer. The two T layers are equivalent for ccp and hep, and for ccp, only P and O layers are interchangeable, and together they are equivalent to the two T layers (considered together). Because of these similarities, ccp, hep, the simple cubic structure, and even bcc structures can be handled in the PTOT system. It also applies to much more complex structures. The PTOT system provides a framework for considering the mechanism of formation and transformation of crystal structures. The transformations of structures of metals, ccp, hep, and bcc, are of particular interest. These are considered in detail in Chapter 4. 

We have discussed structures of O2, S6, and Ss- Now we consider Se, Te, and Po. Six crystalline forms of selenium have been reported ot-Se (stable under normal conditions), a-monoclinic, and (3-monoclinic Se, and three cubic forms deposited as thin films by vacuum evaporation. Metallic a-Se is trigonal, but also described as a distorted simple cubic structure, 3PO, similar to the structure of Te with more distortion for Se. There are infinitive helices parallel to the c axis. The space groups of a-Se are Dg, P3i21 and Dg, P3221 for the other enantiomorph (see Section 10.1.3). 

When the r+/r. ration is between 0.732 and 0.999 the structure is cubic. In this case the cations and anions are not that different is size and the structure corresponds to the simple cubic structure (sc) that we heard about for metals (section 2.3.2 Lattice structures). Anions and cations will be placed in a simple cubic structure so that each cation will be surrounded by eight anions and vice versa. A unit cell for such a structure is show in... 

Another way to envision crystal strnctnres, especially useful in metals where bonds are not nsnally directional, is to think of the atoms as stacking in layers, much as fruit is stacked at the grocery store. For example, the simple cubic structure can be envisioned as one layer of atoms arranged in a sqnare pattern with the next layer stacking directly over the first, so that the atoms in one layer align exactly on top of the atoms in the layer beneath it, as shown here ... 

Face-centered cubic (fee) Again, this is a simple cubic structure, but now, with an additional atom at the center of each square face (Figure 2.1b). Most late transition and noble metals adopt this structure as well as the inert gas solids and some of the alkaline earth elements ... 

Metals A and B form an alloy or solid solution. To take a hypothetical case, suppose that the structure is simple cubic, so that each interior atom has six nearest neighbors and each surface atom has five. A particular alloy has a bulk mole fraction XA = 0.50, the side of the unit cell is 4.0 A, and the energies of vaporization Ea and Eb are 30 and 35 kcal/mol for the respective pure metals. The A—A bond energy is aa and the B—B bond energy is bb assume that ab = j( aa + bb)- Calculate the surface energy as a function of surface composition. What should the surface composition be at 0 K In what direction should it change on heaf)pg, and why ... 

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