Section D Structure and Bonding in Solids

1、SiO2 crystals(long range order) silicate glasses(short-range order) While anatase is not an equilibrium phase of TiO2, it is kinetically stabilized. At temperatures between 550 and about 1000 C, anatase transforms to the equilibrium rutile phase, increasing its specific gravity to 4.2. The t e m p e

2、 r a t u r e o f t h i s transformation strongly depends on the impurities or dopants present in the material as well as on the morphology of the sample。 Rutile has among the highest refractive indices at visible wavelengths of any known crystal, and also exhibits a particularly large birefringence

3、and high dispersion. Owing to these properties, it is useful for the manufacture of certain optical elements, especially polarization optics, for longer visible and infrared wavelengths up to about 4.5m. Ti atoms are gray; O atoms are red The titanium cations have a coordination number of 6 meaning

4、they are surrounded by an octahedron of 6 oxygen atoms. The oxygen anions have a co-ordination number of 3 resulting in a trigonal planar co-ordination The brookite structure is built up of distorted octahedra with a titanium ion at the center and oxygen ions at each of the six vertices. Each octahe

5、dron shares three edges with adjoining octahedra, forming an orthorhombic structure photocatalytic activity Any atom at a corner of a unit cell is shared between eight cells, any at an edge between four, and any on a face between two.(c) 2003 Brooks/Cole Publishing / Thomson Learning Determining the

6、 Number of Lattice Points in Cubic Crystal SystemsDetermine the number of lattice points per cell in the cubic crystal systems. If there is only one atom located at each lattice point, calculate the number of atoms per unit cell.Example 3.1 SOLUTIONIn the SC unit cell: lattice point / unit cell = (8

7、 corners)1/8 = 1In BCC unit cells: lattice point / unit cell = (8 corners)1/8 + (1 center)(1) = 2In FCC unit cells: lattice point / unit cell = (8 corners)1/8 + (6 faces)(1/2) = 4The number of atoms per unit cell would be 1, 2, and 4, for the simple cubic, body-centered cubic, and face-centered cubi

8、c, unit cells, respectively.Determining the Relationship between Atomic Radius and Lattice Parameters Determine the relationship between the atomic radius and the lattice parameter in SC, BCC, and FCC structures when one atom is located at each lattice point.The relationships between the atomic radi

9、us and the Lattice parameter in cubic systemswe find that atoms touch along the edge of the cube in SC structures.ra20In BCC structures, atoms touch along the body diagonal. There are two atomic radii from the center atom and one atomic radius from each of the corner atoms on the body diagonal, so34

10、0raIn FCC structures, atoms touch along the face diagonal of the cube. There are four atomic radii along this lengthtwo radii from the face-centered atom and one radius from each corner, so:240raCalculating the Packing FactorThe packing factor is the fraction of space occupied by atoms, assuming tha

11、t atoms are hard spheres sized so that they touched their closest neighbor. The general expression for the packing factor is :cellunit of volume)atomeach of )(volumeatoms/cell of(number Factor PackingCalculating the Packing FactorCalculate the packing factor for the FCC cell.SOLUTIONIn a FCC cell, t

12、here are four lattice points per cell; if there is one atom per lattice point, there are also four atoms per cell. The volume of one atom is 4r3/3 and the volume of the unit cell is 30a74. 018)2/4()34(4)( Factor Packing24r/ cells,unit FCCfor Since,)34)(atoms/cell (4 Factor Packing330303rrraaDetermin

13、ing the Density of BCC IronDetermine the density of BCC iron, which has a lattice parameter of 0.2866 nm.Example 3.4 SOLUTIONAtoms/cell = 2, a0 = 0.2866 nm = 2.866 10-8 cmAtomic mass = 55.847 g/molVolume of unit cell = = (2.866 10-8 cm)3 = 23.54 10-24 cm3/cellAvogadros number NA = 6.02 1023 atoms/mo

14、l32324/882. 7)1002. 6)(1054.23()847.55)(2(number) sadrocell)(Avogunit of (volumeiron) of mass )(atomicatoms/cell of(number Density cmgFrenkel defects in crystal (interstitials）Schottky defects in crystal (vacancies） hcp structure ccp or fcc structure ABABAB ABCABCABC If the spheres are in contact bo

15、th structure give 74% filling of space by the spheres, with the remaining 26% outside them The hcp structureOne unit cell of the fcc structurebcc structureThe body-centered cubic (bcc) has fairly efficient space filling with the value 68%. Each atom has eight near-neighbors, but there are six others

16、 slightly further away.The fourteen types of Bravais lattices grouped in seven crystal systems.Cesium chloride (AB) 8:8Rocksalt (AB) 6:6Rutile (AB2) 6:3Fluorite (AB2) 8:4Znic blend (AB) 4:4Rhenium trioxide (AB2) 4:2The CdI2 structure can be imagined as a hexagonal close packing of iodine, with 1/2 o

17、f the octahedral holes filled by cadmium. This structure can be generated from the NiAs structure by removing half of the cations in alternating layers. 6:3Nickel arsenide (AB) 6:6)g(Cl)g(Na) s (NaCl)Cl(A)Na( I)Cl(B21)Na(H)NaCl(H)NaCl(H2atfLThe enthalpy of formation of NaClThe enthalpy of atomization of Na solidThe bond enthalpy of Cl2The ionization energy of NaThe electron affinity of Cl2)g(Cl)g(Na)g(Cl)g(Na)Cl