Ch12 sample exercise

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Ch12 sample exercise

  1. 1. Sample Exercise 12.1 Qualitative Comparison of Semiconductor Band Gaps Will GaP have a larger or smaller band gap than ZnS? Will it have a larger or smaller band gap than GaN? Solution Analyze: The size of the band gap depends upon the vertical and horizontal positions of the elements in the periodic table. The band gap will increase when either of the following conditions is met: (1) The elements are located higher up in the periodic table, where enhanced orbital overlap leads to a larger splitting between bonding and antibonding orbitals: or (2) The horizontal separation between the elements increases, which leads to an increase in the electronegativity difference and the bond polarity. Plan: We must look at the periodic table and compare the relative positions of the elements in each case. Solve: Gallium is in the fourth period and group 3A. Its electron configuration is [Ar]3 d 10 4 s 2 4 p 1 . Phosphorus is in the third period and group 5A. Its electron configuration is [Ne]3 s 2 3 p 3 . Zinc and sulfur are in the same periods as gallium and phosphorus, respectively. However, zinc, in group 2B, is one element to the left of gallium and sulfur in group 5A, is one element to the right of phosphorus. Thus we would expect the electronegativity difference to be larger for ZnS, which should result in ZnS having a larger band gap than GaP. For both GaP and GaN the more electropositive element is gallium. So we need only compare the positions of the more electronegative elements, P and N. Nitrogen is located above phosphorus in group 5A. Therefore, based on increased orbital overlap, we would expect GaN to have a larger band gap than GaP. Additionally, nitrogen is more electronegative than phosphorus, which also should result in a larger band gap for GaP. Check: Looking in Table 12.2 we see that the band gap of GaP is 2.26 eV. ZnS and GaN are not listed in the table, but external references show the band gaps for these compounds to be 3.6 eV for ZnS and 3.4 eV for GaN.
  2. 2. Sample Exercise 12.1 Qualitative Comparison of Semiconductor Band Gaps Will ZnSe have a larger or smaller band gap than ZnS? Answer: Because zinc is common to both compounds and selenium is below sulfur in the periodic table, the band gap of ZnSe will be smaller than ZnS. Practice Exercise
  3. 3. Sample Exercise 12.2 Identifying Types of Semiconductors Which of the following elements, if doped into silicon, would yield an n-type semiconductor? Ga; As; C. Suggest an element that could be used to dope silicon to yield a p-type material. Answer: Because Si is in group 4A, we need to pick an element in group 3A. Boron and aluminum are both good choices—both are in group 3A. In the semiconductor industry boron and aluminum are commonly used dopants for silicon. Practice Exercise Solution Analyze: An n-type semiconductor means that the dopant atoms must have more valence electrons than the host material. Silicon is the host material in this case. Plan: We must look at the periodic table and determine the number of valence electrons associated with Si, Ga, As, and C. The elements with more valence electrons than silicon are the ones that will produce an n-type material upon doping. Solve: Si is in column 4A, and so has four valence electrons. Ga is in column 3A, and so has three valence electrons. As is in column 5A, and so has five valence electrons; C is in column 4A, and so has four valence electrons. Therefore, As, if doped into silicon, would yield an n-type semiconductor.
  4. 4. Sample Exercise 12.3 Designing a LED Green light-emitting diodes can be made from a solid solution of GaP and AlP. These two compounds have band gaps that are 2.26 eV and 2.43 eV, respectively. If we assume that the band gap of a Ga 1- x Al x P solid solution varies linearly from GaP to AlP, what composition would be needed for the emitted light to have a wavelength of 520 nm? Solution Analyze: The wavelength of the emitted light has an energy that is nearly equal to the band gap of the semiconductor. The band gap depends upon the composition. Plan: First we must convert the desired wavelength of emitted light, 520 nm, to an energy in eV. Next we must estimate the value of x that gives a band gap of this size. Solve: The wavelength of light can be determined from Equation 6.3: The band gap changes linearly from 2.26 eV for GaP to 2.43 for AlP. Therefore, we can estimate the band gap of any composition in the Ga 1- x Al x P solid solution from the following expression:
  5. 5. Sample Exercise 12.3 Designing a LED If we assume that the band gap of an In 1- x Ga x N solid solution varies linearly from InN (2.4 eV) to GaN (3.4 eV), what composition would be needed for the emitted light to have a wavelength of 410 nm? Answer: The emitted light has an energy of 3.02 eV (485 × 10 –19 J). Therefore, the composition of the semiconductor should be In 0.37 Ga 0.63 N. Practice Exercise Solution (continued) We can solve for x by rearranging the equation and inserting the desired value of the band gap: Thus, the desired composition would be Ga 0.29 Al 0.71 P.
  6. 6. Sample Exercise 12.4 The Stoichiometry of Cross-Linking If we assume four sulfur atoms per cross-link connection, what mass of sulfur per gram of isoprene, C 5 H 8 , is required to establish a cross-link as illustrated in Figure 12.28 with 5% of the isoprene units in rubber? How would you expect the properties of vulcanized rubber to vary as the percentage of sulfur increases? Explain. Answer: The rubber would be harder and less flexible as the percentage of sulfur increases because of an increased degree of cross-linking, which covalently bonds the polymer chains together. Practice Exercise Solution Analyze: We are asked to calculate the mass of sulfur required per gram of isoprene. Plan: We need to evaluate the ratio of sulfur atoms to isoprene units based on Figure 12.15, then scale the required mass of sulfur to take account of the 5% cross-linking. Solve: We can see from the figure that each cross-link involves eight sulfur atoms for every two isoprene units; this means that the ratio of S to C 5 H 8 is four. Thus, with 5% (0.05) of the isoprene units cross-linked, we have
  7. 7. Sample Exercise 12.5 Properties of Liquid Crystals Solution Analyze: We have three molecules of differing molecular structure, and we are asked to determine whether any of them is likely to be a liquid crystalline substance. Plan: We need to identify the structural features of each case that might induce liquid crystalline behavior. Solve: Molecule (i) is not likely to be liquid crystalline because it does not have a long axial structure. Molecule (iii) is ionic; the generally high melting points of ionic materials and the absence of a characteristic long axis make it unlikely that this substance will exhibit liquid crystalline behavior. Molecule (ii) possesses the characteristic long axis and the kinds of structural features that are often seen in liquid crystals (Figure 12.39). Which of the following substances is most likely to exhibit liquid crystalline behavior?
  8. 8. Sample Exercise 12.5 Properties of Liquid Crystals Suggest a reason why the following molecule, decane, does not exhibit liquid crystalline behavior: CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 3 Answer: Because rotation can occur about carbon–carbon single bonds, molecules whose backbone consists predominantly of C—C single bonds are too flexible; the molecules tend to coil in random ways and thus are not rodlike. Practice Exercise
  9. 9. Sample Integrative Exercise Putting Concepts Together Solution Analyze: For part (a), we need to recall what we have learned about sp , sp 2 , and sp 3 hybridization and geometry. (Section 9.5) For part (b), we need to write a balanced equation. For part (c), we need to use the ideal-gas equation. (Section 10.4) For part (d), we need to recall the definitions of endothermic and exothermic, and how bond enthalpies can be used to predict overall reaction enthalpies. (Section 8.8) A conducting polymer is a polymer that can conduct electricity. Some polymers can be made semiconducting; others can be nearly metallic. Polyacetylene is an example of a polymer that is a semiconductor. It can also be doped to increase its conductivity. Polyacetylene is made from acetylene in a reaction that looks simple but is actually tricky to do: (a) What is the hybridization of the carbon atoms, and the geometry around those atoms, in acetylene and in polyacetylene? (b) Write a balanced equation to make polyacetylene from acetylene. (c) Acetylene is a gas at room temperature and pressure (298 K, 1.00 atm). How many grams of polyacetylene can you make from a 5.00-L vessel of acetylene gas at room temperature and room pressure? Assume acetylene behaves ideally, and that the polymerization reaction occurs with 100% yield. (d) Using the average bond enthalpies in Table 8.4, predict whether the formation of polyacetylene from acetylene is endothermic or exothermic.
  10. 10. Sample Integrative Exercise Putting Concepts Together Solution (continued) Plan: For part (a), we should draw out the chemical structures of the reactant and product. For part (b), we need to make sure the equation is properly balanced. For part (c), we need to convert from liters of gas to moles of gas, using the ideal-gas equation ( PV = nRT ); then we need to convert from moles of acetylene gas to moles of polyacetylene using the answer from part (b); then, we can convert to grams of polyacetylene. For part (d), we need to recall that Δ H rxn = ∑ (bond enthalpies of bonds broken) – ∑ (bond enthalpies of bonds formed). Solve: (a) Carbon always forms four bonds. Thus, each C atom must have a single bond to H and a triple bond to the other C atom in acetylene. As a result, each C atom has two electron domains and must be sp hybridized. This sp hybridization also means that the H—C—C angles in acetylene are 180°, and the molecule is linear. We can write out the partial structure of polyacetylene as follows: Each carbon is identical but now has three bonding electron domains that surround it. Therefore, the hybridization of each carbon atom is sp 2 , and each carbon has local trigonal planar geometry, with 120° angles. (b) We can write:
  11. 11. Sample Integrative Exercise Putting Concepts Together Solution (continued) Note that all atoms originally present in acetylene end up in the polyacetylene product. (c) We can use the ideal-gas equation as follows: Acetylene has a molar mass of 26.0 g/mol; therefore, the mass of 0.204 mol is Because Δ H is a negative number, the reaction releases heat, and is exothermic.

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