Prof. Thomas Greenbowe Prof. Gordon Miller Dr. Cristina Bonaccorsi This exam consists of 2 parts on 7 pages CHEM 177 Hour Exam IV November 12, 2012 Name KEY Recitation TA Recitation Section Parts Page 3 Page 4 Page 5 Page 6 SUBTOTAL Grand Total Grading Points 24 pts 21 pts 31 pts 24 pts 55 pts 100 pts Score on Scantron sheet TA Name Sections Time Berry 14, 38 10, 2 Bobbitt 36, 48 2, 3 Boschen 3, 9 8, 9 Burgin 53, 58 4, 5 Carraher 6, 12 9, 10 Castle 40, 46 2, 3 Chaudhary 47 3 Cole 19 11 De Silva 21, 31 11, 1 lood 54 5 riedrichsen 7, 30 9, 1 Ganesh 5, 11 8, 9 Geraskin 20, 26 11, 12 Geraskina 17, 23 10, 11 Hochstein 35, 42 1, 2 Joshi 16, 29 10, 12 Keller 1 8 Khan 22, 49 11, 4 Kumar 28, 34 12, 1 Kwolek 4, 10 8, 9 Lei 44, 52 3, 4 Lesoine 15, 57 10, 5 Nelson 33, 41 1, 2 Nguyen 24 12 Peeraphatdit 32, 39, 45 1, 2, 3 Thooft 43, 51 3, 4 Vangaveti 18, 27 11, 12 Wang 50, 55 4, 5 Wanninayake 2, 8 8, 9 Weerasekara 13, 25 10, 12 Xu 37, 56 2, 5 NOTE: To receive full credit for problems on Pages 5-6, make sure to draw clearly all bonds (single, double and/or triple), lone pair(s), and charges. Replacement Exam (OPTIONAL): December 3 rd (same times/locations as other hour exams). This exam is all multiple choice questions. In recitation (November 15), you will indicate to your TA which exam you are trying to replace; you must confirm this selection via Blackboard no later than NOON, Tuesday, November 27. inal Exam: December 13 th (4:30-6:30pm). NOTE: There may be different room assignments for the inal Exam than for the Hour Exams.
Please read the following instructions carefully before proceeding! Part I of your exam will be computer graded. In order for the computer to identify who you are, it is important that you complete the information section properly. You must use a #2 pencil and completely fill in the appropriate circles on the BLUE computer scan sheet. 1. 2. To help you code the correct circles, first write your last name, first name, and middle initial in the boxes (skip a space between each). Then darken the circles that match the letters in the box above it. See the sample to the right. Write the middle nine digits of your ISU identification number in the boxes A-I. Do not skip any spaces. Below each number, darken the circle that matches this number. or example, 123456789. See the sample at bottom right. 3. Write your recitation section number in the special code area, boxes K-L, as two digits. Do not skip any spaces. or example, if you are in section 8, write 08. Again, darken the circle that matches the number above it. See the sample at bottom far right. In Part I, select the one best answer for each question. Place your answer on the computer answer sheet by darkening the proper circle for that question. Your computer scan sheet will be your official answer sheet for Part I. All material (exam, answer sheet, scratch paper) must be returned to your TA in order for us to grade your exam.
Part I. (45 pts) Multiple Choice. (15 questions at 3 pts each). 3 1. Which of the following correctly represents the second ionization of calcium? a) Ca + (g) Ca 2+ (g) + e b) Ca (g) Ca 2+ (g) + 2e c) Ca (g) + e Ca 2 (g) d) Ca + (g) + e Ca 2+ (g) e) Ca + (g) + e Ca (g) 2. Of the following elements, has the least negative electron affinity. a) K b) Ca c) As d) Se e) Br 3. A tin atom has 50 electrons. Electrons in the shell experience the lowest effective nuclear charge. a) irst (n = 1) b) Second (n = 2) c) Third (n = 3) d) ourth (n = 4) e) ifth (n = 5) 4. Given the electronegativities below, which covalent single bond is most polar? Element: H C N O Electronegativity: 2.1 2.5 3.0 3.5 a) C H b) N H c) O H d) O C e) O N 5. The oxidation number of sulfur in SO 2 is a) 0 b) +2 c) +3 d) +4 e) +6 6. Among the three bonds C N, C=N, and C N, the C N bond is a) strongest/shortest b) strongest/longest c) weakest/shortest d) weakest/longest e) intermediate in both strength and length 7. Among the following molecules, only is polar. a) BeCl 2 b) CBr 4 c) N 3 d) AlCl 3 e) Cl 2 8. The total number of bonds in the H C C C C C C C N molecule is a) 4 b) 6 c) 8 d) 12 e) 15 Total Pts
4 9. In ethylene (H 2 C=CH 2 ), the C=C double bond results from overlap of orbitals and overlap of orbitals on the C atoms. a), sp,, p z b), sp 2,, p z c), sp 2,, sp 2 d), sp 2,, p z e), sp 3,, s 10. According to molecular orbital theory, overlap of two s atomic orbitals produces a) one bonding molecular orbital and one hybrid orbital b) two bonding molecular orbitals c) two bonding molecular orbitals and two antibonding molecular orbitals d) two bonding molecular orbitals and one antibonding molecular orbital e) one bonding molecular orbital and one antibonding molecular orbital 11. Which drawing represents a bonding molecular orbital for a homonuclear diatomic molecule? a) b) c) d) 12. Which of the following molecular species is stabilized by resonance? a) CO 2 b) N 3 c) BH 3 d) O 2 2 e) HCN Use the following MO diagram for questions (13) and (14). 2p * π 2p * π 2p 2p 2s * 2s 13. Which of the following molecular species has the lowest bond order? a) NO b) O 2 + c) CO d) NO 14. Which of the following molecular species will be paramagnetic? a) NO b) N 2 c) 2 d) NO + 15. The formal charge on the N atom in the correct Lewis structure of nitrosyl trifluoride, NO 3, is a) 0 b) +1 c) +2 d) +3 e) +5 Total Pts
PART II. (31 points) Lewis Structures. 16. (21 pts total) Consider the nitrite ion, NO 2. (a) (6 pts) Draw all relevant Lewis structures for the nitrite ion, NO 2, taking into account resonance and the octet rule). Be sure to show all lone pairs. # valence electrons = (5) + 2(6) + (1) = 18 e (9 pairs) 5 (b) (3 pts) or each atom in the nitrite ion, label (clearly) the formal charge (underneath the atom symbol in your Lewis structures). (c) (2 pts) What is the oxidation state of N in the nitrite ion? Ans. +3 (d) (4 pts) What is the molecular structure (shape) of the nitrite ion? Briefly explain your choice. BENT, there are 3 electron pair domains at N (1 lone pair, 1 single bond, 1 double bond). The lone pair repels the two bond-pair domains to create a bent-shaped molecule. (e) (2 pts) What is the average bond order of the N O bond in the nitrite ion? Explain your choice. Both N-O bonds are equivalent: (2 + 1) / 2 = 3/2 Ans. 3/2 = 1.5 (f) (4 pts) In the nitrous acid molecule, HNO 2, the hydrogen atom could be bonded to either the N atom or an O atom. Using your Lewis structures in part (a) above, draw a valid Lewis structure of HNO 2. In HNO 2, the H would behave like H + and, so, would be more attracted to O, the more electronegative element in nitrite (O also shows a 1 formal charge). In this Lewis structure, the formal charges of all atoms are now 0. 17. (10 pts total) Use this skeleton of the molecule glycine, H 2 NCH 2 COOH, for questions (a)-(c): (a) (6 pts) Complete the Lewis structure of glycine, clearly marking all multiple bond(s) and lone pair(s). (b) (2 pts) What is the hybridization at the N atom? Tetrahedral (c) (2 pts) Estimate the O C O bond angle. (sp 2 at C) sp 3 120 Total Pts
PART III. (24 pts) Chemical Bonding and Molecular Structures 18. Mixing Ge 4 and Se 4 in a 1:1 molar ratio gives the solid compound with the formula GeSe 8. A controversy exists over whether this compound is (Ge 3 ) + (Se 5 ) or (Se 3 ) + (Ge 5 ). 6 (a) (20 pts) Write Lewis structures and use VSEPR to evaluate the electron pair geometry at the central atom, the molecular structure for the ion, and the hybridization at the central atom. Lewis Structure Electron-Pair Geometry Molecular Ion Geometry Hybridization at Central Atom (Ge 3 ) + (4 + 3 1)/2 = 3e pairs Ge Planar Planar (sp 2 )(p) (Se 3 ) + (6 + 3 1)/2 = 4e pairs Tetrahedral Pyramid (sp 3 ) (Ge 5 ) (4 + 5 + 1)/2 = 5e pairs Ge Bipyramid Bipyramid (sp 2 )(p) (Se 5 ) (6 + 5 + 1)/2 = 6e pairs Se Octahedral Square Pyramid (sp)(p)(p) (b) (4 pts) The observed structure contains a trigonal planar ion. What is the correct description of GeSe 8? (circle one) (Ge 3 ) + (Se 5 ) or (Se 3 ) + (Ge 5 ). Total Pts
7 INORMATION Bond Order (Diatomics) = (# electrons in Bonding MOs # electrons in Antibonding MOs) / 2 ormal Charge = (# valence electrons in neutral atom) (# electrons assigned from Lewis structure) Important VSEPR Descriptions: Linear, Planar, Tetrahedral, Bipyramidal, Octahedral Electronegativities of the Main Group Elements Rules for Oxidation States of Elements in Compounds (1) 0 in elements; (2) Alkali metals: +1 (3) Alkaline-earth metals: +2 (4) luorine: 1 (5) Oxygen: 2, except for peroxides (O 2 2 : 1) (6) Halogens (Cl, Br, I): 1, except when combined with or O (7) Hydrogen: +1 with nonmetals; 1 with metals (8) Sum of oxidation states in molecule/ion = total charge 1 A 1 1 H 1.0 1 3 Li 6.9 4 1 1 Na 2 3.0 1 9 K 3 9.1 3 7 Rb 8 5.5 5 5 Cs 1 3 3 8 7 r (2 2 3 ) 2 A 2 4 Be 9.0 1 1 2 Mg 2 4.3 2 0 Ca 4 0.1 3 8 Sr 8 7.6 5 6 Ba 1 3 7 8 8 Ra 2 2 6 3 B 3 2 1 Sc 4 5.0 3 9 Y 8 8.9 5 7 La 1 3 9 8 9 Ac 2 2 7 4 B 4 2 2 Ti 4 7.9 4 0 Zr 9 1.2 7 2 Hf 1 7 8 1 0 4 Rf (2 6 1 ) 5 B 5 2 3 V 5 0.9 4 1 Nb 9 2.9 7 3 Ta 1 8 1 1 0 5 Db (2 6 2 ) 6 B 6 2 4 Cr 5 2.0 4 2 Mo 9 5.9 7 4 W 1 8 4 1 0 6 Sg (2 6 3 ) Pe riodic Table of t he Ele me nt s 7 B 7 2 5 Mn 5 4.9 4 3 Tc (9 8 ) 7 5 Re 1 8 6 1 0 7 Bh (2 6 2 ) 8 2 6 e 5 5.8 4 4 Ru 1 0 1 7 6 Os 1 9 0 1 0 8 Hs (2 6 5 ) 8 B 9 2 7 Co 5 8.9 4 5 Rh 1 0 3 7 7 Ir 1 9 2 1 0 9 Mt (2 6 6 ) 1 0 2 8 Ni 5 8.7 4 6 Pd 1 0 6 7 8 Pt 1 9 5 1 1 0 Ds (2 8 1 ) 1 B 1 1 2 9 Cu 6 3.5 4 7 Ag 1 0 8 7 9 Au 1 9 7 2 B 1 2 3 0 Zn 6 5.4 4 8 Cd 1 1 2 8 0 Hg 2 0 1 3 A 1 3 5 B 1 0.8 1 3 Al 2 7.0 3 1 Ga 6 9.7 4 9 In 1 1 5 8 1 Tl 2 0 4 4 A 1 4 6 C 1 2.0 1 4 Si 2 8.1 3 2 Ge 7 2.6 5 0 Sn 1 1 9 8 2 Pb 2 0 7 5 A 1 5 7 N 1 4.0 1 5 P 3 1.0 3 3 As 7 4.9 5 1 Sb 1 2 2 8 3 Bi 2 0 9 6 A 1 6 8 O 1 6.0 1 6 S 3 2.1 3 4 Se 7 9.0 5 2 Te 1 2 8 8 4 Po (2 0 9 ) 7 A 1 7 9 1 9.0 1 7 Cl 3 5.5 3 5 Br 7 9.9 5 3 I 1 2 7 8 5 At (2 1 0 ) 8 A 1 8 2 He 4.0 0 1 0 Ne 2 0.2 1 8 Ar 3 9.9 3 6 Kr 8 3.8 5 4 Xe 1 3 1 8 6 Rn (2 2 2 ) Lant hanides Act inides 5 8 Ce 1 4 0 9 0 Th 2 3 2 5 9 Pr 1 4 1 9 1 Pa 2 3 1 6 0 Nd 1 4 4 9 2 U 2 3 8 6 1 Pm (1 4 5 ) 9 3 Np (2 3 7 ) 6 2 Sm 1 5 0 9 4 Pu (2 4 4 ) 6 3 Eu 1 5 2 9 5 Am (2 4 3 ) 6 4 Gd 1 5 7 9 6 Cm (2 4 7 ) 6 5 Tb 1 5 9 9 7 Bk (2 4 7 ) 6 6 Dy 1 6 2 9 8 Cf (2 5 1 ) 6 7 Ho 1 6 5 9 9 Es (2 5 2 ) 6 8 Er 1 6 7 1 0 0 m (2 5 7 ) 6 9 Tm 1 6 9 1 0 1 Md (2 5 8 ) 7 0 Yb 1 7 3 1 0 2 No (2 5 9 ) 7 1 Lu 1 7 5 1 0 3 Lr (2 6 0 )