Name CHM 4610/5620 Fall 2016 November 15 EXAMINATION TWO SOLUTIONS

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1 Name CHM 4610/5620 Fall 2016 November 15 EXAMINATION TWO SOLUTIONS I II III IV V Total This exam consists of several problems. Rough point values are given to help you judge the value of problems. The total will be scaled to 100 points after the exams are marked. Glance over the entire exam, and then attempt the problems in the order of your choice. You must show your work or explain your reasoning to receive credit for an answer. For calculations, draw a box around your final answer and be sure to include the correct units. Some additional information is provided as a separate handout. You have 55 minutes (class period plus 10%). Good luck! I. (20 points) InSb is one of the major semiconductor materials used for the construction of infrared transmitters and receivers such as those used in remote control devices. A. Determine the wavelength of the electromagnetic radiation that would be emitted from an InSb semiconductor. Reading E off the band gap graph gives around 10 kj/mol. E = 10 kj/mol / x mol -1 = 1.7 x kj x (1000 J /1 kj) = 1.7 x J E = h => = E/h = 1.7 x J / x J s = 2.5 x s -1 c = => = c/ = 3.0 x 10 8 m s -1 / 2.5 x s -1 = 1.2 x 10-5 m Is the wavelength in the IR range, as expected? 1.2 x 10-5 m x (10 9 nm/1 m) = 12,000 nm Longer wavelength than visible ( nm), so that seems reasonable. Strictly speaking, the infrared region picks up at 700 nm and extends to 10 6 nm. B. In order to optimize the conductivity, InSb is often doped with Cd. In the box on the right, sketch a band gap diagram for InSb doped with Cd, and label the valence band, conduction band, and ΔE corresponding to the band gap energy required to achieve conduction. Would this be n-type or p-type doping? see Figure 7.26 in HK&K4 Cd has fewer electrons than In or Sb, so p-type doping energy gap is less, so wavelength would be greater for the doped semiconductor

2 CHM 4610/5620, Exam Two, Fall 2016 page 2 II. (10 points) Before beginning our discussion of molecular orbital theory, we speculated on the best dot structure for C 2. A. Draw your favorite dot structure for C 2. answers vary B. The MO diagrams from our text for diatomic molecules are given on the last two pages of the exam. Using one and only one of these diagrams, populate the MOs with the appropriate number of electrons, and answer the following questions: 1. What is the bond order of C 2? bond order = (8 2)/2 = 3 2. How many unpaired electrons are contained in one C 2 molecule? 0 (zero) 3. How does your dot structure from Part A agree or disagree with your MO diagram? Answer will depend on dot structure from Part A.

3 CHM 4610/5620, Exam Two, Fall 2016 page 3 III. (30 points) The MO diagrams you used in the previous problem are for diatomic molecules formed from elements in the second row of the periodic table. The elements below that row do not usually form double or triple bonds. For example, the most common form of molecular nitrogen is N 2 with an N N triple bond; however, the most common form of molecular phosphorous is P 4 ( white phosphorous ) with P P single bonds. (By the by, Professor Michael Shatruk and his group in our department have been doing some very nice work with another form of phosphorous: The structure of P 4 is shown on the right. It is tetrahedral and belongs to the T d point group. (You can think of the P 4 structure kind of like methane, CH 4, without the central atom.) Each phosphorous atom is connected to the other three atoms by a single bond. A. Determine the reducible representation for all of the vibrations expected for P 4. Do not resolve this representation into its irreducible components. T d E 8C 3 3C 2 6S 4 6 d A x 2 +y 2 +z 2 A E (2z 2 -x 2 -y 2, x 2 -y 2 ) T (R x, R y, R z ) T (x, y, z) (xy, xz, yz) xyz x # stat atoms = total trans rot = vib

4 B. Now determine the reducible representation for all of the stretches expected for P 4. CHM 4610/5620, Exam Two, Fall 2016 page 4 T d E 8C 3 3C 2 6S 4 6 d A x 2 +y 2 +z 2 A E (2z 2 -x 2 -y 2, x 2 -y 2 ) T (R x, R y, R z ) T (x, y, z) (xy, xz, yz) P-P Note: vib = stretches => There are no bending vibrations! The rigid structure prohibits bends without considerable change in the bond lengths. C. Resolve the representation from Part B into its irreversible components. If you re running short on time, you only have to find a couple of the irreducible reps enough for me to see that you know how to resolve the reducible representation. D. For the stretches you found in Part C, which are IR active? And which are Raman active?

5 CHM 4610/5620, Exam Two, Fall 2016 page 5 IV. (10 points) Consider the hypothetical triatomic atom H 3. To keep things simple, let s assume it is bent, so the point group would be C 2v. You can then think of one of the hydrogens as the center atom, bonded to two other H atoms, so we might write the formula as HH 2. A. Construct a qualitative MO diagram for HH 2. Label the MOs as bonding (b), nonbonding (nb), or antibonding (*). Populate the MO diagram with the correct number of electrons. (Before starting, think about which atomic orbitals you need and which you do not need to construct your MO diagram.) B. Sketch all of the MOs from your MO diagram, and label them as you did in Part A. C. Do you expect HH 2 to be stable? Explain. Yes. Two electrons in a bonding MO; none in antibonding MOs. D. What about our assumption that HH 2 is bent, as opposed to linear? Do you think that is a valid assumption? Explain your reasoning. No nonbonding electron density on the central hydrogen atom, so it s probably linear.

6 CHM 4610/5620, Exam Two, Fall 2016 page 6 V. (20) In class we posited that lithium fluoride is arguably the simplest salt, but context is everything A. LiF(s) is an ionic compound, but that characterization is less clear when talking about LiF(g). The bond enthalpy of LiF(g) is reported to be 577 kj/mol, presumably corresponding to H for the reaction LiF(g) Li(g) + F(g) The above reaction is homolytic bond dissociation, splitting the bond evenly to give two neutral atoms. But what about heterolytic bond dissociation to give ionic species in the gas phase, analogous to the ionic species in the lattice for the ionic solid? LiF(g) Li (g) + F (g) Predict which process requires the most energy. heterolytic bond dissociation Explain your reasoning. In general it takes more energy to remove an electron from the metal than you get back giving an electron to a nonmetal. B. Draw and annotate a thermochemical cycle that would allow you to calculate H for the reaction Li(g) + F(g) LiF(g) Li (g) + F (g) Information that may or may not be useful is given in the table on the right. process species H (kj/mol) ionization energy F 1,680 ionization energy Li 520 electron affinity F 328 electron affinity Li 60 bond enthalpy F F 157 bond enthalpy Li Li 110 bond enthalpy Li F 577 IE 1 (Li) EA 1 (F) H o BE(Li F) LiF(g) Li (g) + F (g) Now use your cycle to calculate H. Be sure to show your work. 577 kj/mol kj/mol + ( 328 kj/mol) = 769 kj/mol Does the calculation support your prediction? Explain. Yes (769 kj/mol > 577 kj/mol). The enthalpy required to ionize lithium is greater than the energy you get back giving an electron to fluorine. C. Now consider the reverse reaction: Li (g) + F (g) LiF(g) Is this an acid/base reaction or a redox reaction? acid/base reaction If acid/base, what is then acid and what is the base? Li is the acid and F is the base If redox, what is oxidized and what is reduced?