Homework 1 Solutions - ENGN 0410 Intro to Materials Science Fall 2024 PDF

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This document contains solutions to homework problems in materials science for the Fall 2024 semester. The problems involve determining electron configurations and calculating percent ionic character of chemical bonds.

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ENGN 0410 Intro to Materials Science Fall 2024 Solutions to Homework #1 (100 points total) 1) (12 pt) Without consulting the figures from Callister, determine whether each of the electron configurations given below is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transitio...

ENGN 0410 Intro to Materials Science Fall 2024 Solutions to Homework #1 (100 points total) 1) (12 pt) Without consulting the figures from Callister, determine whether each of the electron configurations given below is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choices. (a) 1s22s22p63s23p63d74s2 (b) 1s22s22p63s23p6 (c) 1s22s22p5 (d) 1s22s22p63s2 (e) 1s22s22p63s23p63d24s2 (f) 1s22s22p63s23p64s1 Solution (a) The 1s22s22p63s23p63d74s2 electron configuration is that of a transition metal because of an incomplete d subshell. (b) The 1s22s22p63s23p6 electron configuration is that of an inert gas because of filled 3s and 3p subshells. (c) The 1s22s22p5 electron configuration is that of a halogen because it is one electron deficient from having a filled L shell. (d) The 1s22s22p63s2 electron configuration is that of an alkaline earth metal because of two s electrons. (e) The 1s22s22p63s23p63d24s2 electron configuration is that of a transition metal because of an incomplete d subshell. (f) The 1s22s22p63s23p64s1 electron configuration is that of an alkali metal because of a single s electron. 2) (10 pt) Compute the percents ionic character of the interatomic bonds for the following compounds: TiO2, ZnTe, CsCl, InSb, and MgCl2. Explain how you got your answer and include the values of the electronegativity that you used for the calculation. Solution The percent ionic character is a function of the electron negativities of the ions XA and XB: − (X A − X B )2 f = (1 − exp( ) X 100 4 For TiO2, XTi = 1.5 and XO = 3.5, and therefore,  2 %IC = 1 − e(− 0.25)(3.5−1.5)  × 100 = 63.2%   For ZnTe, XZn = 1.6 and XTe = 2.1, and therefore,  2 %IC = 1 − e (− 0.25) (2.1−1.6)  × 100 = 6.1%   For CsCl, XCs = 0.7 and XCl = 3.0, and therefore,  2 %IC = 1 − e(− 0.25)(3.0− 0.7)  × 100 = 73.4%   For InSb, XIn = 1.7 and XSb = 1.9, and therefore,  2 %IC = 1 − e(− 0.25)(1.9−1.7)  × 100 = 1.0%   For MgCl2, XMg = 1.2 and XCl = 3.0, and therefore,  2 %IC = 1 − e(− 0.25)(3.0−1.2)  × 100 = 55.5%   3) (14 pt) What type(s) of bonding would be expected for each of the following materials: brass (a copper-zinc alloy), rubber, barium sulfide (BaS), solid xenon, bronze, nylon, and aluminum phosphide (AlP)? Solution For brass, the bonding is metallic since it is a metal alloy. For rubber, the bonding is covalent with some van der Waals. (Rubber is composed primarily of carbon and hydrogen atoms.) For BaS, the bonding is predominantly ionic (but with some covalent character) on the basis of the relative positions of Ba and S in the periodic table. For solid xenon, the bonding is van der Waals since xenon is an inert gas. For bronze, the bonding is metallic since it is a metal alloy (composed of copper and tin). For nylon, the bonding is covalent with perhaps some van der Waals. (Nylon is composed primarily of carbon and hydrogen.) For AlP the bonding is predominantly covalent (but with some ionic character) on the basis of the relative positions of Al and P in the periodic table. 4) (10 pt) In metallic bonding, the outermost electrons go into states that are not localized around a particular atom. Therefore, they are free to move across the solid in response to an applied electric field, i.e., conduct electricity. In materials with ionic bonding, the electron is bonded tightly to the atom so it is not free to conduct electricity. 5) (10 pt) a) In CO2 the bonding electrons reside between the atoms with no electron density available to bond to anything else. Therefore it forms self-contained molecules. When Si and O bond, there are bonds formed in each of the tetrahedral directions so that the Si and O can form a 3-dimensional network. b) The CO2 molecules bond to each other weakly via secondary bonds (dipole interactions). This is called a molecular solid. Because the molecule does not have a permanent dipole, the bonding is Van der Waals. Because the bond is weak, the sublimation temperature is low. 6) (10 pt) a) To calculate the interatomic spacing, find the value of r where the energy is minimized: ∂E (r )  2  ao  2 9  ao 9  =α   −    = 0 ∂r  r  r  r  r   This occurs when the value of r = a0 (9/2)1/7 = 1.24 a0. b) The bond energy equals the value of the energy at the equilibrium spacing. Evaluating E at 1.24 a0 yields a value of -0.506 α c) To determine the stiffness or spring constant of each material, calculate ∂2E/∂r2 at the minimum value (when r= 1.24 a0). ∂ 2 E (r )  a 2  ao 9   5.92α  = α − 6 4  + 90 11  o  = ∂r 2  r  r   ao2   When the bond is compressed by 5 % (i.e., Δr = 0.05 (1.24 a0) ) then the energy changes by ½ ∂2E/∂r2 (Δr) 2 = 0.0114 α. 7) (12 pts) a) The interatomic potentials have the shapes shown below. 0.5 0 0 1 2 3 potential E1 energy -0.5 potential E2 -1 -1.5 r (angs) b) From the shape of the curves it is clear that potential 2 has a greater curvature at the minimum of the curve. Therefore, the stiffness of material 2 is greater. c) The potential that is more asymmetric will have the higher thermal expansion coefficient. Therefore material 1 will have the greater thermal expansion coefficient. 8) (12 pts) a) Atoms that are touching (r=a0) have the bond energy E0 = -1. Atoms that are at spacing of r=√2 a0 (across the diagonal of the square) have bond energy E1 = -0.799. Atoms that are at the far ends of the rhombus have a spacing of r=√3 a0 and a bond energy of E2 = -0.658 The rhombus has 5 bonds of energy -1 and 1 bonds of energy -0.658. The total energy of the configuration is -5.658. The square configuration has 4 bonds of energy -1 and 2 bonds of energy -0.799. The total energy of the configuration is -5.6. The tetrahedral configuration has 6 bonds of energy -1 for a total energy of -6. The tetrahedron is therefore the lowest energy configuration. b) The significance for solids is that the lowest energy corresponds to the densest packing of the spheres. 9) (10 pts) a) For each atom in the octahedral configuration, there are 4 atoms that are touching it (distance r=a0) and 1 atom at a distance r=√2 a0. Counting the bonds in the diagram, there are 12 with energy E0 = -1 and 3 with energy E1 = -0.799. Therefore the total energy is equal to 12E0 + 3E1 =-14.4 and the energy per atoms is equal to -14.4/6 = - 2.4. b) In problem 8, the energy per atom for the tetrahedral configuration is equal to -1.5. This shows that the energy per atom gets lower as the cluster gets bigger which means that larger clusters are more stable than small clusters. This means that individual atoms can lower their bond energy by forming a solid.

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