Activity 5&6: Metals and Hexagonal Close-Packing

Transcription

1 Chemistry 150 Name(s): Activity 5&6: Metals and Hexagonal Close-Packing Metals are chemicals characterized by high thermal and electrical conductivity, malleability and ductility. Atoms are the smallest unit of matter that retains chemical properties. A large number of metal atoms gives rise to the properties of metals listed above, but what makes some atoms metallic and others not? The answer lies in the distribution of subatomic particles that make up each atom. Subatomic particles are the constituents of atoms; there are three principal subatomic particles of interest: protons and neutrons (both of which exist in the nucleus of the atom) and electrons (which orbit the nucleus). Subatomic particles have a small amount of mass and two of them, the proton and the electron, also carry an electrical charge. By convention, the charge on a proton is positive and the charge on an electron is negative. Thus, a proton is said to be attracted to an electron; and two electrons repel each other. It just so happens that the charge on one proton exactly neutralizes the charge on one electron. Since metal atoms are all neutral, therefore the number of protons they contain equals the number of electrons they contain. But just because they are neutral does not mean they don t interact! Note that a collection of atoms with no interatomic interactions would fall apart; there would be no metal items! Electrons provide the interaction between atoms that hold them together. This interaction is called bonding. In the case where electrons are shared between two neighboring atoms, the type of bond is called covalent. In metals, the electrons around a particular atom don t particularly care about what other atom they re being shared with. Therefore, the bonds in metals are non-directional or delocalized covalent bonds. In fact, the electrons around metal atoms are so delocalized that they can leave their original atom and quite happily exist around another metal atom, then another, then another. The model for electrons in metals is thus called a sea of electrons with metal nuclei spaced regularly throughout the sea. 1. How does the sea of electrons model help explain the electrical conductivity of metals? Hint: Just what is electricity anyway? 2. a. On the atomic level, how does heat manifest itself?

2 b. So how does the sea of electrons model help explain the thermal conductivity of metals? We have six ICE kits, so form six groups; obtain an Institute for Chemical Education (ICE) Solid-State Model Kit. Take out the plastic tub containing the clear and green spheres; remove the green spheres (they will roll away, so put them someplace contained). The clear spheres will represent metal atoms, with the nucleus represented by the center of the tube through the sphere. Bunch up the clear spheres against the side of the tub such that they form a two- or threelayer deep structure; remove your hand. 3. How does this demonstration illustrate how the sea of electrons model explains the malleability and ductility of metals? When the atoms favor no particular bonding orientations (such as in a metal), the crystalline structure can be conveniently described using the hard sphere model. Metallic bonding being non-directional in nature, the structure of most elemental metals is accurately described by a dense packing of hard spheres. In this model, hard spheres (representing nearest neighbor atoms) touch one another to form a densely packed structure. In the case of elemental metals, all the atoms are identical. Three relatively simple crystal structures are found for most metals: face-centered cubic (FCC), hexagonal close-packed (HCP) and body centered cubic (BCC). FCC and HCP correspond to very dense atomic arrangements called a closed-packed structure, while BCC is a less dense structure. The diagram below illustrates the lattice pattern of these different packing arrangements. We have already discussed FCC and BCC in lecture. The purpose of this part of the lab is to construct hard-sphere crystal models in order to see the connection

3 between FCC cubic unit cells and cubic close-packed (CCP) hexagonal unit cells. As you may recall from lecture, FCC and CCP are the same. 4. Look in the Instruction Manual and build the FCC model (page 27) and the FCC (body diagonal) model (page 28). How do they compare? 5. Turn to the CCP (body diagonal) on page 26 and compare the layer diagrams of CCP to FCC (body diagonal). The CCP layer diagrams show only the atoms in one cubic unit cell (colored spheres, outlined numbers). How do they compare? Since FCC and CCP are supposed to be the same, they should have the same cell volume to atom ratio. In lecture, the volume of the FCC structure was shown to be 16 2r 3 and to contain four atoms, so that the volume per atom is 4 2r 3. Show that the CCP hexagonal unit cell (page 25) has the same volume per atom. Note that a regular hexagon can be divided into equilateral triangles and that the 3 height of an equilateral triangle is times the length of a side. The volume of a 2 hexagonal prism is the area of the hexagonal base time the height of the prism. The height of the CCP prism is the same as the length of the body diagonal of the FCC unit cell contained within it, which is 3 times the length of the FCC unit cell side, or 2 6r. Another number you ll need to know is how many spheres each unit cell contains entirely within the unit cell s boundaries. Come up with a method for counting the atoms in a hexagonal unit cell this will not be exactly the same as a cubic cell! 6. Figure out the number of spheres entirely within the boundaries of the CCP hexagonal unit cell: 7. Determine the volume of the CCP hexagonal unit cell. 8. Determine the volume per atom for the CCP hexagonal cell.

4 Build a model of HCP (page 24) and apply what you have learned about hexagonal cells to answer the following questions. 9. Figure out the number of spheres entirely within the boundaries of the HCP hexagonal unit cell: 10. Determine the volume of the HCP hexagonal unit cell. Hint: the height of this unit cell differs from that of CCP. 11. Determine the volume per atom for the HCP hexagonal cell. 12. What can you say about the packing efficiencies of HCP, CCP and FCC? Now build the following structures and fill in the table:

5 Molybdenite, MoS 2, p. 75 Nickel Arsenide, p. 35 Zinc Blende, p. 49 Wurtzite, p Fill in the table Model Which color ball represents which element? Number of atoms per unit cell (list each element) Empirical formula Which crystal system? Molybdenite, MoS 2 (p. 75) Nickel Arsenide (p. 35) Zinc Blende (p. 49) Wurtzite (p. 55) 14. Complete the following table of packing efficiencies, based on your understanding of HCP in question 12

6 SC FCC/CCP BCC HCP 52% 74% 68% The density of a particular metal will depend on how efficiently each individual atom is packed within the structure. 15. The densest metals would tend to have which structure(s), all other factors being equal? At present, why certain metals pack with certain structures is not known. 16. Examine the table below. These elements exist next to each other on the periodic table so they have approximately the same atomic mass (their atoms weigh nearly the same). The atomic radius is the size of their spheres measured in nanometers ( = m, so they are, of course, quite small). That information is included to show you that they are all about the same size. The following numbers belong in the last column: 8.9, 8.9, 7.9. Enter them, according to your rule in question 15, in the cells in the last column. Metal Crystal structure Atomic radius (nm) Density (g/cm 3 ) Iron BCC Cobalt HCP Nickel FCC