Name: 1: /33 Grade: /100 2: /33 3: /33 +1 free point. Midterm Exam I. CHEM 181: Introduction to Chemical Principles September 20, 2012 Answer Key

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1 ame: 1: /33 Grade: /100 2: /33 3: /33 +1 free point Directions: Do all three problems. Midterm Exam I EM 181: Introduction to hemical Principles eptember 20, 2012 Answer Key how all of your work neatly and clearly. Do not skip steps. Partial credit will be awarded for all problems. orrect answers will not receive credit if your work is not shown. If you are not sure exactly what a question means, ask! ot all problems are of equal difficulty, but all are worth the same fraction of the overall grade. 1

2 1. Draw Lewis electron structures for the following molecules and ions. Do not exceed an octet on any atom. Mark all non-zero formal charges. Include all reasonable resonance structures. Mark which structures are, and mark major and minor structures. You can either omit extremely minor structures, or mark them as extremely minor. (a) 3 (onnect major the to an oxygen; avoidminor bonds) v. minor (b) 2 (ne, the, and the are connected to the.) minor major v.v. minor major minor Drawing the other isomer (with one each attached to and ) and its resonance structure is fine, too. 2

3 (c) v. minor v. minor minor major v.v. minor (d) 2 2 (four-membered ring, alternating,,, ) minor major v.v. minor 3

4 2. The first through sixth ionization energies of carbon, in attojoules (1 aj = J) are: Z Element I 1 I 2 I 3 I 4 I 5 I In photoelectron spectroscopy, a single photon detaches an electron from an atom A: A + photon A + + e (with kinetic energy) (a) arbon has 1s, 2s, and 2p electrons. Which of these can be detached using light with wavelength λ = 100 nm? Explain your answer. First calculate the photon energy: E = hc λ = J s m s m = J = 1.99 aj This is enough to pull of the first electron, which is in a 2p orbital. The 2s orbital is lower in energy than the 2p orbital; the question is, how much lower? It s not going to take a full 7.67 aj to pull off an electron, as that s the energy needed to remove a 2s electron after both 2p electrons are gone. n the other hand, you can eyeball the large jump between I 2 and I 3, compared to the much smaller jumps between I 1 and I 2, and I 3 and I 4, and guess that the 2s orbital is approximately (or at least) 1 aj more stable than the 2p. o 100 nm can only remove a 2p electron, as the 2s is too strongly bound. 4

5 (b) What is the maximum speed for an electron detached with 100 nm light? The kinetic energy released is KE = hc λ E binding = 1.99 aj 1.81 aj = 0.18 aj Plugging this in with the mass of the electron: KE = 1 2 m ev 2 2E v = m e = 18 J kg = J 11 kg = J kg 1 kg m2 s 2 1 J = m s 1 (c) Which electrons (1s, 2s, and/or 2p) can be detached using 28 nm light? Explain your answer. You can do another calculation, or go ahead and say that this light has 100/28 times the energy: aj = 7.09 aj Based on the previous discussion, this will be enough to detach either a 2s or 2p electron. It is almost enough to detach a 2s electron even after both 2p electrons are gone. The 1s orbital energy is going to be less than 62.8 aj, but it s not going to be too much less: the 2s and 2p electrons don t shield the 1s electrons from the nucleus much. 5

6 (d) Assume no energy is lost as heat or remains in the atom; that is, all excess energy goes into kinetic energy. If light with a wavelength of 2 nm is used, how many different electron speeds will be observed? (Explain your answer.) The energy is 50 times higher than the 1.99 aj you calculated in part (a), so almost 100 aj: plenty of energy to knock out any one of the electrons. The possible processes are: (1s 2 2s 2 2p 2 ) + photon + (1s 2 2s 2 2p 1 ) + energy (1s 2 2s 2 2p 2 ) + photon + (1s 2 2s 1 2p 2 ) + energy (1s 2 2s 2 2p 2 ) + photon + (1s 1 2s 2 2p 2 ) + energy There are 3 possible electron speeds. (2p is removed) (2s is removed) (1s is removed) 6

7 v. minor 3. The cyclopentadienyl anion, 5 5, has a five-membered ring of carbons, with one hydrogen bonded minor to each carbon. major v.v. minor (a) Draw the Lewis electron structure, including all resonance structures, for cyclopentadienyl. There are five resonance structures, all. (The molecule is aromatic.) (b) What is the bond order, and what is the average formal charge on each carbon atom? For any given bond you pick, it will be double-bonded in two of the five resonance structures, and single-bonded in the other three. That s seven bonds total over five resonance structures: bond order = 7 5 All five carbon-carbon bonds are identical in this molecule. The formal charge, by a similar calculation, is 1/5. 7

(c) The molecular orbitals for cyclopentadienyl look like haracterize each orbital as bonding, antibonding, or non-bonding, and write this (or b, ab, nb or the like) next to each of the above

8 (c) The molecular orbitals for cyclopentadienyl look like haracterize each orbital as bonding, antibonding, or non-bonding, and write this (or b, ab, nb or the like) next to each of the above orbitals. A, B, and D are bonding, and and E are antibonding. (d) Rank the molecular orbitals in terms of energy, lowest to highest. (If two orbitals are very close in energy, do not worry overmuch about which one you list first.) lowest energy A < B = D < = E highest energy It is not obvious that B and D should be exactly the same energy, but they are. Any answer that has B and D as higher in energy than A but lower than and E is marked as correct; ditto any answer with and E as the two highest-energy orbitals. (e) Which orbitals are filled? What does M theory predict for the bond order of this molecule? There are 26 electrons total. 10 of these are taken up in the 5 sigma bonds, and another 10 in the 5 sigma bonds. There are 6 electrons left for our π molecular orbitals, shown above. All of these electrons go into bonding orbitals (A, B, and D), so they contribute an additional 3 bonds around the ring. This suggest 8 bonds total (5 σ, 3 π), and a bond order of 8/5. 8

Final Exam. CHEM 181: Introduction to Chemical Principles Answer Key. 1. The major resonance structure of 1,4-diazepine is shown below:

Final Exam. CHEM 181: Introduction to Chemical Principles Answer Key. 1. The major resonance structure of 1,4-diazepine is shown below: Final Exam EM 181: Introduction to hemical Principles Answer Key 1. The major resonance structure of 1,4-diazepine is shown below: or Show all reasonable resonance structures; label them as major (if equivalent