Exam 3_Answer_Key_FINAL

Exam 3_Answer_Key_FINAL - LAST NAME FIRST NAME(Please...

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Unformatted text preview: ___________ LAST NAME, FIRST NAME (Please PRINT) C102A Examination III ANSWERS Fall, 2009 G. N. Lewis (1875–1946) Honor Code: I pledge on my honor that I have neither given nor received improper aid on this examination. _____________Linus Pauling________ (Signature) Directions: 1. No books or notes may be used during this test. There should be no data stored on your calcula ­ tor for use on the exam. This includes, but is not limited to, periodic tables and chemical equa ­ tions and formulas. 2. There are a total of 8 sections (some with several parts) for a sum of 100 points. 3 Be certain that, wherever applicable, you show your work. Problems in sections 7 ­8 that are an ­ swered without showing work will not receive credit. 4. Please note: there are no intentionally misleading questions on this test. Each problem should be taken at its face value. You must, however, be certain to read each problem carefully! 5. If you have any questions please ask your professor. “The strength of science lies in its naïveté… if we could foresee all the obstacles that lie in our path we would not attack even the first, but would settle down to self centered contemplation.” —Gilbert N. Lewis, in The Anatomy of Science (1926) DO NOT REMOVE THIS PAGE FROM YOUR EXAM!!!!! Please check the box by the section in which you are officially enrolled: Prof. List (MWF, 8:10 ­9:00) Prof. Hanusa (MWF, 9:10 ­10:00) Prof. Lukehart (TTh, 8:10 ­9:25) Prof. Phillips (MWF, 11:10 ­12:00) Exam Score________ 2 SECTION 1 (10 PTS. TOTAL; 2 PTS. EACH) Circle either True or False for each answer. (1) Both tetrahedral and trigonal pyramidal molecules have four VSEPR electron groups about the central atom. (2) The octet rule will always apply to covalent compounds as long as there are no more than 4 atoms around the central atom. (3) Electronegativity is a concept that is useful in deciding how many electrons are involved in a bond. (4) Compounds with atoms that are more polarizable display stronger intermolecular forces than do those with less polarizable atoms. (5) Molecules containing a central atom with sp3d hybridization will all have a trigonal bipyramidal molecular geometry. True True True True True False False False False False SECTION 2 (27 PTS. TOTAL; 3 PTS. EACH) Pick the one best answer; if you circle more than one, you will receive NO credit! (1) Analysis of an unknown substance showed that it has a high boiling point and is brittle. It is an insulator as a solid but conducts electricity when melted. Which of the following substances would have those characteristics? a. HCl b. Al c. KBr d. SiF4 e. I2 (2) The lattice energy of MgCl2 is the energy change for which one of the following processes? a. Mg(s) + Cl2(g) → MgCl2(s) b. Mg(g) + 2Cl(g) → MgCl2(s) c. Mg2+(s) + 2Cl–(g) → MgCl2(g) d. Mg2+(g) + 2Cl–(g) → MgCl2(s) e. MgCl2(aq) → MgCl2(s) (3) Arrange aluminum, nitrogen, phosphorus and indium in order of increasing electro ­ negativity. a. Al < In < N < P b. Al < In < P < N c. In < Al < P < N d. In < P < Al < N e. None of the above orders is correct. 3 (4) Choose from the list below the organic molecule with a carbon atom having the same hybridization as that of nitrogen in NOCl (N is the central atom) a) CH4 b) C2H4 c) C2H2 d) CO2 e) none of the above (5) The solute ­solvent interactions when NaCl dissolves in water are primarily of the type: a) dipole ­dipole. b) ion ­dipole. c) dipole ­induced dipole. d) ion ­ion. e) none of the above. (6) Select the compound with the most negative lattice energy. a) CaS(s) b) BaO(s) c) NaI(s) d) LiBr(s) e) MgO(s) (7) Given the mean bond enthalpies for C ­O (360 kJ/mol), C=O (743 kJ/mol), and C≡O (1077 kJ/mol), predict the bond enthalpy for the CO bond in the formate anion, [HCO2]– (carbon is the central atom). a) 360 kJ/mol b) 552 kJ/mol c) 743 kJ/mol d) 910 kJ/mol e) 1077 kJ/mol (8) Which of the following should have a dipole moment greater than zero? a) PF3 b) AlCl3 c) CO32– d) CH3+ e) none of these would have a dipole moment greater than zero (9) One of the following statements about molecular orbitals is always true—which is it? “A molecular orbital…” a) has increased electron density in the region between the bonded atoms b) can accommodate the total number of electrons in the valence shell of the atoms c) is formed from the atomic orbitals of the less electronegative atom d) can accommodate a maximum of two electrons e) defines the spectroscopic characteristics of the atoms forming a chemical bond 4 SECTION 3 (10 PTS. TOTAL; 2 PTS. EACH) In the center column, write “>”, “<”, “=”, or “?” (can’t tell) to describe the relationship be ­ tween the following quantities. The first one has been done for you. Number of Chem 102A instructors < < > > Number of Chem 102A students The angle around Br in the ion [BrF2]– (Br is central) Bond order of H2– Boiling point of trans ­difluoro ­2 ­butene, CH3 CC H3C F F (1) The angle around S in the molecule NSF (S is central) (2) Bond order of H2. (3) Boiling point of cis ­difluoro ­2 ­butene, H3C CH3 CC F F (4) Polarity of P–Cl bond (5) Boiling point of NH3 > > Polarity of S–Cl bond Boiling point of PH3 5 SECTION 4 (10 PTS. TOTAL; 2 PTS. EACH) Match the following with the best possible answer. (1) A molecule that violates the octet rule is ___E______. (2) A nonpolar molecule containing polar bonds is ___C______. (3) The hybridization around the central carbon of allene, C C C is ___L______. H H H H (4) H ­C ­H bond angle in CH3 ___V______. − (5) A compound that contains both ionic and covalent bonds ___J______. (A) O3 (B) AsF3 (C) CCl4 (D) PCl3 (E) SeF4 (F) CH2 (G) PF3 (H) SO2 (I) (J) NaCl KCN (M) sp2 (N) sp3 (O) sp3d (P) sp3d2 (Q) SbF3 (R) (S) (T) 109.5° 180° 120° (U) 90° (V) <109.5° (K) NH3 (L) sp 6 SECTION 5 (12 PTS. TOTAL) Complete the following table. Chemical formula ICl4+ ClF5 Cl Lewis dot struc ­ ture (2 pts each) I Cl Cl Cl F F F Cl F F Electron group geometry (ar ­ rangement) (1 pt each) Type of hybrids orbital that the central atom forms (1 pt each) Molecular geometry (1 pt each) Does this mole ­ cule have a di ­ pole moment greater than zero? (1 pt each) trigonal bipyramidal octahedral sp3d sp3d2 see-saw square (-based) pyramid yes yes 7 SECTION 6 (13 Pt.) NF’ SAID The exotic compound fluorine azide (FN3) is explosive when condensed, but it can be (care ­ fully!) decomposed to yield the reactive product nitrogen monofluoride, NF. A molecular orbital diagram for NF is given at right below: F (a) Complete the following table. The lowest energy orbital has been done for you. E1 E2 Orbital # of elec ­ Name of or ­ label trons bital σ1s A 2 σ∗1s D B 2 C D E1 F 2 2 1 0 σ2s σ2p π∗2p σ∗2p C (b) Bond order of the NF molecule: ____2_____ (c) Is the NF molecule paramagnetic? ___yes____ B A (d) Predict whether the FN cation (NF+) will have a shorter or longer bond than the NF molecule. (Circle one) Bond in NF+ will be longer / shorter than in NF 8 SECTION 7 (8 PTS.) Credit will not be given if detailed step by step work is not shown. Historically, one of the most common uses of Born ­Haber cycles was to calculate electron affinities, as they are difficult to measure experimentally. Consider the formation of CuCl2 (∆Hf° = –186 kJ mol–1). Use any of the following information to determine the electron affinity of chlorine. The heat of sublimation of Cu(s) is 338 kJ/mol; the first ionization energy of Cu(g) is 746 kJ/mol; the second ionization energy of Cu(g) is 1948 kJ/mol; the electron affinity of cop ­ per is 118 kJ/mol; the bond enthalpy in Cl2(g) molecule is 242 kJ/mol; the first ionization energy of Cl(g) is 1251 kJ/mol; lattice energy of CuCl2(s) is –2772 kJ/mol. Cu(s) → Cu(g) C u (g ) → C u + (g ) + e − ∆Hsub = 338 kJ/mol IE1 = 746 kJ/mol − Cu+(g) → Cu2+(g) + e Cl2(g) → 2Cl(g) IE2 = 1948 kJ/mol ∆Hbond broken = 242 kJ/mol 2Cl(g) + 2e → 2Cl (g) − − 2EA = 2x kJ/mol ∆Hlattice = −2772 kJ/mol ∆Hf = −186 kJ/mol Cu2+(g) + 2Cl (g) → CuCl2 (s) − Cu(s) + Cl2(g) → CuCl2 (s) ∆Hf = ∆Hsub + IE1 + IE2 + ∆Hbond broken + 2EA + ∆Hlattice = −186 kJ/mol 2EA = 2x = –688 kJ/mol EA = x = –344 kJ/mol ____–344_____ kJ/mol 9 SECTION 8 (10 PTS.) Credit will not be given if detailed step by step work is not shown. a) Consider the following two molecules, both with empirical formula C4H6: H CC HCCH [1] H H H → H CH CCC [2] H H H H Using the bond enthalpy table at right, compute the enthalpy change for the change of molecule [1] to molecule [2]. Bonds broken: 3 C–C 1 C=C Bonds formed: 2 C=C 1 C–C ∆Hrxn = (3 C–C + C=C) – (2 C=C + 1 C–C) = (2 C–C) – C=C = 2(347) – 615 = +79 Bond Bond Enthalpy [kJ/mol] C–H 414 C–C 347 C=C 615 O=O 495 C=O 730 O–H 464 C≡C 811 ____+79_____ kJ/mol b) Using the bond enthalpy table above, compute the enthalpy change in the following reac ­ tion: [2] + (11/2)O2 → 4 CO2 + 3 H2O Bonds broken: 1 C–C Bonds formed: 8 C=O 2 C=C 6 O–H 6 C–H 11/2 O=O ∆Hrxn = (1 C–C + 2 C=C + 6 C–H + 11/2 O=O) – (8 C=O + 6 O–H) = [347 + 2(615) + 6(414) + 11/2(495)] – [8(730) + 6(464)] = 6783.5 – 8624 = –1841 ____–1841 _____ kJ/mol ...
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This note was uploaded on 03/17/2010 for the course CHEM chem102 taught by Professor Adamlist during the Spring '10 term at Vanderbilt.

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