AK10 - 4. (E) For simplicity, the 3 lone pairs on halogens...

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Unformatted text preview: 4. (E) For simplicity, the 3 lone pairs on halogens are not shown in the structures below. Add 6 electrons per halogen atom. . (a) (b) I (F6) .0 no , g : : o C_ s : H '0 \ _“ o. u l l C— .. I-I—C—C—H / I F H (d) (8) Cl \ u S O H—C=C—H .0 Cl - 7 fl ‘ 22. (M) a . r (a) HZNOH is favored because same charges don’t reside adjacent to each other.“ ’ H H 1:3 \ oo o. _ \ o. N_—o——H (5'—N———H ll / / H H H ,‘l FC(N) = 5—2—3 : 0 . . ‘ FC(N) = 5—2—2 = +1 ‘ FC(O) = 6-2-2 = +2 FC(O) = 6-2-3 = +1 it (b) S=C=S is favored, because FC on everything is zero. on o. \ to no S:C=S C SZS ' ii} 00 on o- o- FC(C) = 4—0—4 = o FC(C) 2 4-4-2 = -2 a; FC(S) = 6—4-2 = 0 ‘ FC(=S=) = 6--0-4 = +2; FC(=S) = 6-4-2 = 0 36. (c) FANZO is favored. o. ' .. no on :N F o : : F 1:1. 3 00 no on .. (d) OSC12 is favored. Even though the formal charges of all elements on both OSC12 and OClgS are zero, OSC]; represents a much more realistic structure because the O and Cl atoms are in their normal bonding configuration. 51.. O (31'. no 0 :2. 2 ... 0 0 /oo. 0 0 0 /s : . .. I .. / .00 O 2:8 : O=CI: 61.. °'. .. o. no. 0 gl. 0 0 (e) F38N is favored. no 0 \:F: 0 7 +1 :'F': 0 .' n 00\ I 00 :1:0 ITI 1.3.: is ' 1:10 0 -1 26. (E) The total number of valence electrons is (2 x 7) + (2 x 6) = 26 valence electrons, or 13 pairs of valence electrons. It is unlikely to hav atom; that would require an expanded octet on F. The most plausible structure is: = i5——'S' Isl—i5 : (M) (a) Group 14 (b) Group 16 except oxygen o. u I _— ._— oo '0 '9.‘_C_'.O. 3 [ 0 N——-o: .0 .. o. (c) Group 17 except fluorine (d) Group 13 00 2“ o. o _ :o: :o: :o‘: .0. l I. O. .. Q. .o ———-s-—o H :o-— 81—0 81—0: on | 0' o. l 00 | on :o: :3: :3: e F as a central 41. (M) C1-|—->F . ~30 §=E=Wx47ffi=all4atowardsF. d 162.8x10’ In 1.602x10 C Rb-l—’F -3(.1 *1 a = 1i: W“”Dx——le—J¥ = 0.785e, towards F. d 227.0x10‘ In 1.602x10 C r Sn+-—->O um _ 5 = :5 = %— = 0.4918 , towards 0. d 183.3-><10l m - 1.602x10 C Ba-l—FO , —3[1 ‘ ' 6=fi=wc.m/Dx—l%.:0.855e,towardso. d 194.0x10 m 1.602x10 C '52. (M) (a) Structures A and B are equivalent and most important. . A B C .0 :03 0 1 :3: 4 33: '1 00 oo- 50 I o. 0' I. H—9. 5 £1: «1—»- H—g—S 2 +—I- H 0:3 0/,“ ,u I +1 u -1 .0: a : :0: (b) Structure A is the most important. A B H H . | 0 0 l 0 .. 0 :T—CEN: eke—*- +1 TZC N —1 O. H H (c) Both structures are equivalent. ‘ A B — 0 :0: ‘ -1 :0: ll .. *1 ___.. F——-C O: H F—C—O I. C. (d) All structures are equivalent. A B O. O. O. 0. +1 S N: -1 S N I H I l ,etc... ' . O O 0 H .8: 0 a}: ..0 ,etc... 56. g; (D) In N02, there are 5 +(2 x 6) = 17 valence electrons, 8 electron pairs and a lone electron. N is the central atom and it carries the lone electron. A plausible Lewis structure is (A) which has a formal charge of 1+ on N and l— on the single—bonded 0. Another Lewis structure, with zero formal charge on each atom, is (B). The major difference is that one of the oxygen atoms carries the lone electron. In both cases, due to the unpaired .electron we expect N02 to be paramagnetic. We would expect a bond to form between two NO2 molecules as a result of the pairing of the lone unpaired electrons in the N02 molecules. If the sec'ond Lewis structure for N02 is used, the one with zero formal charge on each atom and the lone electron on oxygen, a plausible structure for N204 is (C). If the first Lewis structure for N02 where there are formal charges and a lone electron on ' nitrogen .is used to form a bond between molecules, a plausible Lewis structure containing a N —N bond results (D). Resonance structures can be drawn for this second version of N204. A N7 N bond is Observed experimentally in N204. In either structure, the product N204, has all electrons paired up in bonding and non-bonding lone pairs; the molecule is expected to be (and is) diamagnetic. Lewis structures for (A) a (D) are shown below. :9: (A) :Q—ng: (B) (9.. G3 _ . =9~I~i=9r ‘ '9tN-9' 919*?” =9—1r=0: 6” @ .,:.._k..~.._..:.. _.J grm '99_§*0 (M) ||-| l I ll H—p—p—H + [El—El —+ H—(lz—cngl + H—EI H H H H ' Analysis of the Lewis structures of products and reactants indicates that a C — H bond and a Cl — Cl bond are broken, and a C — Cl and a H —— Cl bond are formed. kJ kJ Energy required to break bonds = C——H+ Clw—Cl == 414 wk] +243 — = 65 7 —— mol mol mol - . . 1 k] k] k] Energy realized by forming bonds = C—Cl + Hi-(sl = 339 —+ 431 —— = 770 — I mol mol mol AH=657 kJ/molfi770 kJ/mol=—ll3kJ/mol ' it. _. F: C) SF?" F L‘s/21 :‘F: .F H‘s, F/FNF c’l 1+ 06/ 0c» 09‘ '{Tifioncd Ptfiovnldqni \ZJ: O ./ Q's: "{fimhed‘fd d\ (30%- :é—Cl\g't'5:’ + fim \fl ech‘cd #25.) 03 CA F2 ' —-—- 1 F '«e F lmfaf F 0) mar _-§._.Ab;_¢v. F 3““ on) F—g'fii—F fl Ot/Jmhed ml ‘0') C" F3 air?» ELF—4; -. .. ‘ . =53 F \ -'”(_\-——F \ F +—S\nqped :E: F F | _,F .‘C‘. F, u-NF. c) m.) E F SH F ~69? ch MN r file; F x? ; I0 i: F’g’mq: F’ 1.‘ “gemu /‘: F / " \ F F see—Sum Stfiuuve phmar Pas? T POL“ a (“,11I1‘.F_____CK \ I ' 9 ., q ‘1 0/? MIC" <_\ I c1 0‘ b b». mm'dQ‘ *“3 ‘ ‘ \ w +e’nahedmi Pd; 0 \ “d (9 u m ——--'P'-—"~ 0‘ /\ a O Odbhe (1 HM 5. Provide the best 3-dimenslonal Lewis structures of the following molecules. if you are lousy at drawing, you can add a word to help us understand your picture (e.g., H20 is “bent”). & a. 8e03, b. BrF4' Q fill . E) . . a f3 2 \ 131-! "“‘" d.- F If; {7' 0 a d’" \ O /. . 6 a.) o r c. Xe02F4 d. TeOF4 F' f E? O 3 t F . l' “‘3 Ell - w r F~><Q-F (“T4230 rx| F / H | F .i e IO:> f. SO; - 6 4" c G) I :1- “ ~‘ s 4:“ *0 c" 2:9" 0‘59 O " I/ O O / o it) ‘ o o 6’ Q Q) J 9 52032 h CzNz ( r“) {5 fr urea-4:13;), ‘ Q9 1,") \ 3 ° " co :5 \ a C) “ / Q‘ <3 <5 I 0 O C) L") aw") Q i The bond distance in HCl (9) is 1.274 A and its dipole moment is u = 1.05 D. Calculate the partial charges on the H and the Cl. 5 1:771 a 0:3??3X1‘U.90C9+w ’ “- ...
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This note was uploaded on 06/10/2011 for the course CHEM 2090 taught by Professor Zax,d during the Spring '07 term at Cornell.

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AK10 - 4. (E) For simplicity, the 3 lone pairs on halogens...

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