fall midterm 1 - 5.1- 14A2F01-1 Baur CHEMISTRY 14A '...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 5.1- 14A2F01-1 Baur CHEMISTRY 14A ' LECTURE SECTION 2 FALL QUARTER, 2001 ‘ First Midmafion - ‘ ‘ Friday, October‘19; 2001 Name CWQ TA: (check one) (LAST) (FIRST) Damian Aherne Kim Miller Mollie Cavanaugh Will Molenkamp_ Section Number Natalie Gassman - QUESTION N0. CREDIT SCORE _ 2 _ s — 4 - — | N ST R CT I O N S 1. There are _four questions. Answer each in the space prgvidgd; you may write on the backs of sheets if you need more room, but no loose sheets of paper will be accepted. . Your work must be clearly W and legible; credit will not be given for unintelligible reSROnses. V . Method or reasoning must be shown for‘all computational items; no credit for merelyputting down an answer, even it correct. , . ' Make sure that all answers have the appropriate number of significant figures. Give urlitsfor all answers. unless dimensionless. . _ 1 Open book examination. Any written or printed materials may be used. . _ DO NOT LIFT THIS COVER SHEET AND BEGIN WORK UNTIL INSTRUCTED TO DO $0!- on» N939"? th fII win at ifn d H 1.008 O 16.00 Na 22.99 Cl 35.45 Cr. 52.00 Br 79.90 He 4.003 F 19.00 Mg 24.30 Ar 39.95 Fe 55.85 Ag 107.87 C 12.01 Ne 20.18 .3 32.08 K 39.10 CU 63.55 Cd112.41 N 14.01 ‘ Ca 40.08 . 'Zn 65.39 | 126.90 NA (Avogadro's Number) = 6.022 x 1023 V I Photon energy (Planck Law): a = hv = hc/7x. c(velocity of light) = 3.00 x 108 ms1 _ ' h(PIanck’s Constant) = 6.63 x 10-34 J5 Bohr formula: En = -2.178 x 10-18 J (22/n2) 1 e.v.(electron volt) = 1.60 x 10-19J Coulomb energy between two charges Q1 and 02;: E3 = 2.31 x 10-19 J nm(Q1Qzlr). (Q. in multiples of the elementary Charge) The periodic table and alpgiablg are on the back of this sheet. CHECK TO MAKE SURE THAT THIS EXAMINATION PAPER HAS FOUR QUESTIONS AND FIVE PAGES , -2- Name 1. (30 points) Consider the reaction between iodine pentoxide and sulfur tetrafluoride: I205 +, 5SF4 —> 2|F5 + 5OSF2 . (a) (15 points) Is this a redox process? To decide, assign oxidation numbers to each atom in each of the chemicals cies involved 1 oint each) pe 1205 ( p Iii 0:3 SF4 S +5: F411 IFS l +5 F -_l OSF2 o_—_z, S_+_‘i F:‘_’ Is any atom oxidized or reduced? If so, say which, and by how many electrons. (6 points) #0 mm m amp/22.1,} 016 6530453), 77,95 ,5 ddiE£DJK Fences; (b) (15 points) Suppose 350.0 grams each of IZO5 and SF4 are mixed and allowed to react according to the equation above. -Which is the limiting reagent? (8 points) 4’11 = 337% 2 39‘07 __ /' 413 MM ’05 (Jar/22 ¢ 5x/é.o Him" 33‘} yam-l 0 :: ' '= 359.07 _~_ 3. mo; (32,1 + 4X Mo) 7444:.“ 1083/ 7 Mot-r @QUIze SmoLs 5th pee law; $7,051 $0 muff? a Mom 3&4 awaits; nzmrwl-m MM ML 1:05 :5 .——— X/,0¢¢ML5-;0§ Mole'or H" : 7 ' 5/. patfiasgolf éérfiét 5isforme5<1?S?7goi ts)I5 Lou/77mm J 39"“qu 1” SF Slit/65 $5,; /5L/»1/T‘/A/é mu; IF; F005; /5 (omfipgi’mm/To-F 4/_ l .. ' " - 22 ML M455 17‘5- : /.30 MOI/5 x (/27 + Madam» -/,30mLs x2 7 I : 287?. -3- 14A2F01-1 Name 2. (20 points) (a) (5 points) Calculate the frequency in 3-1 (Hz) of photons with wavelength 4.20 x 10-7 m (420 nm). ’ __ C 3 60 '0 5-: V— -—— : __'___i‘_’._.’_“___. ___ I HIS—l A ‘7’.zox/o~7 M 7’ six/0 (b) (5 points) Calculate the energy of a single 4.20 x 10-7 m photon, in J and in e.v. (electron volts). 9‘ RV -= C'UX’O-” {F5 x7./Ll—></0"‘.‘S“' :— 7¢X/0”7T -,7 /,60x/U"‘7 T(€’.lf-)" (c ) (10 points) When a photon of wavelength 4.20 x 10-7 m strikes the surface of a particular photoelectric substance, an electron with kinetic energy 1.38 x 10-19 J is emitted. Find the work function W of the, substance, in J, and the maximum wavelength which a photon could have and still cause the photoelectric effect to occur. -I I £3 w + Exam» KIA/£T’C Ell/5.6%)“ :w+ figs/WU 9J- W‘ £-/: BMW/7T: 4.74x/o"7J-/,35/x/0 a _ A c kaIMUM a/A—VfiLEA/kw- [A]: {Ml/V ’ AMAX A ; KC :(éJSx/OJVTE' )(BMWO‘BMr' > M” W . 3-36 x10"? :r -7“ = 59mm M - {w 41% '—— -4- 1 4A2 F01 -1 Name 3. (20 points) (a) (6 points) For each of the following ground state electronic configurations of neutral atoms. give the number of W electrons, according to Hund’s Rule. . - mflgumtign W' A. (He) 2s22pa 3 B. (He) 2522p5 / o. (Kr)‘ 5514d8 - V ' 3 0. (Ne) 3s23p6 ‘ O From this set. the atom wtih the largestfirst ioni ation energy i345“? "If D the atom with thejargestelectron affinity is y. ‘ , V ~ g. ' (b) (14 points) Below are five arrangements of electrons among valence orbitals. For each, you are to select all species from the list to the rig t which correspond to it (in some cases more than one species fits), and specify whether the configuration corresponds to the ground state or an excited state. Assume that all inner shell orbitals (which are not shown) are filled. .mMMm EmMMM mMm LMMME ‘ W W - . .. C', N, 0, 0+: F, F+, Ne, .LL.—I.L_ 6‘ L______ 3&2... Ne+. P-,-s+, Cr. cm, 1; 3p 3p 3p . ‘ Co, Co+3, Pd, cd+2 35 iuu_h ifl:;_iflflE2 :L 2p 2p 2p ‘ S 3— Cn (0+3 UA/ A6712ro «NW 0'» gag-515315515 4s -—————' .—————G‘"’ ? ‘ exam? co”) itl Jfli£;;@flfl_ é; 2p 2p 2p ‘ S .LL‘iiiLlLL E1; 4d4d4d 4d 14A2FO1-‘l Name 4. (30 points) ' (a) (8 points) As discussed in lecture, the skeletal geometry of the triatomic ion l3- is linear resulting from trigonal bipyramidal orbital geometry of class AX2E3. There is also a triatomic ion l3+; predict the skeletal geometry of this ion, first finding the number of valence electrons, their orbital geometry and class designation (AXmEn). Describe the orbital geometry in words (briefly) and give a Lewis diagram of the ion including lone pairs. 2 o "f. , * Number of valence electrons . Lewis diagram: ‘7 ‘ The orbital geometry is 75779 “9me , l The class designation is A XL E 7. ‘- It, The skeletal geometry is BEA/T C 33-5 JA K5 (+2.0) . (b) (8 points) Does the molecule ClgO, which has 20 valence electrons, have Cl as the central atom (similar to structure of N20) or as the cdentral atom (similar to structure of H20)? Answer this by drawing a Lewis diagram mm for each possibillity and evaluating the formal charges which would be present for each. . Ql Cl Q M l \l | I q I l ‘ . a i \ \ I sued—0: .C/Qw—cQ‘ 526.41on: 0 +‘ '" o O O ' ' 777% m IDPLE ‘ 77m flee/Wkng WITH 0 w . Conclusion: #195 375/30 Awe/"44 Wfifis) HEIVCE /5 77759 0&7— d4/E _ A (c) (14 points) Use the set of diatomic molecule molecular orbitals discussed in lecture and given on p. 148 of the text (Figure 3.34 for species with less than 12 electrons, Figure 3.35 for species with 12 or more electrons) to tell which of the following second row neutral diatomic molecules is most stabke abd which is Least stable by finding the bond order for each; also give the number of unpired electrons in each: ~ §D§£l§§ W I ‘ Emmme LiC 56‘ EA 0“ )7140“ "9 WWW? ‘ é / ‘ — L w". 1 *) 5 CF He . (arm/10’ )L &:¢)(z{¥,.)(2fisfl‘)( V711 1 / - US 48' {urngmdUL '0 0 _ L z 7.. 'l l ‘NF [18 (m)Z(“mz?z'v/1)aflwhpwflfyflmflx) ’4‘ ’2 Most stable C F ' Least stable L} B ...
View Full Document

Page1 / 5

fall midterm 1 - 5.1- 14A2F01-1 Baur CHEMISTRY 14A '...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online