E 45 - Fall 2010 - Gronsky - Midterm 1 (solution)

Ν ε x ε z e45 fall 10 midterm 01 solutions

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ν = ε x ε z E45 Fall 10 Midterm 01 Solutions Professor R. Gronsky page4 of 11
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2. Bonding c . When r = R (not possible in ionic solids), the radius ratio cited in ( b ) above is r/R = 1, and the coordination number is 12, as found in such metallic crystals as Al, Cu, Ni, Ti, and many others. However, when carbon atoms bond to other carbon atoms ( r = R ) forming diamond, the coordination number is 4. Explain , citing concepts of covalent bonding. Covalent bonding is described as the "mutual" sharing of electrons, facilitated by the overlap of valence electron orbitals. Unlike metallic bonding, in which those orbitals are fully delocalized, covalent bonding results from localized orbital overlap, rendering the bonds "directional." The bonds form along directions of greatest orbital overlap . In the case of diamond, carbon bonds covalently after sp 3 hybridization, which limits the number of C bonds that can be formed by localized orbital overlap to 4, resulting in CN = 4, and the "diamond cubic" crystal structure shown in the figure. d . Carbon-carbon bonds in polyethylene are also covalent, yet polyethylene softens at 120°C while diamond softens (melts) at 3500°C. Explain . Polyethylene is a polymer with a backbone chain of covalently-bonded carbon atoms, but the chains are bonded to one another by electrostatic dipole interactions. Such dipole interactions are induced by the large groups of atoms on adjacent backbone chains, which generate weak bonds, called "secondary" bonds, to distinguish them from "primary" bonds that form between individual atoms or ions. In diamond, ALL bonds between all carbon atoms are primary (covalent) bonds, resulting in a much stronger material, as the difference in "softening" temperatures between these two materials confirms. ! " # C H H H H C n n C H H H H C E45 Fall 10 Midterm 01 Solutions Professor R. Gronsky page5 of 11
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3. Lattice Geometry a . Identify the family to which the following planes in a cubic lattice belong. Use Miller index notation. Referring to the plane closest to the origin in this figure, its intercepts with the coordinate axes are , , . Taking the reciprocals, clearing fractions (unnecessary here) and enclosing in parentheses, the answer is (008) Check : the other planes are parallel to this one, and appear at intersections ½ and along the z axis, confirming that they belong to the same (008) family. b . The figure below shows one member of the (111) family of planes in a face-centered cubic structure. Specify in Miller index notation the lattice directions connecting the following pairs of atoms. Check : Dot product = 0 for all directions in the (111) plane with [111] normal. ! " # ! " # " # $ % −→ OA = [ ¯ 101] −−→ OB = [ ¯ 211] −−→ OC = [ ¯ 110] −−→ AB = [ ¯ 110] −→ AC = [ ¯ 12 ¯ 1] E45 Fall 10 Midterm 01 Solutions Professor R. Gronsky page6 of 11
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3. Lattice Geometry c . Specify in Miller-Bravais notation the family of planes containing the following planes as members.
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