cem181f06e2 - Chemistry 181E Film @006 EXAM] II I...

Info iconThis preview shows pages 1–9. 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

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

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

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

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chemistry 181E Film @006 EXAM] II I fiursflai 9 NOV M6 WRITE your Namd Student Numbfir Begifiatign Timei on ax Ihfirg am 5 questionfi QIl this exam, Mill] 20 pointsl bash. mug your gliswer2 showing alfl world Fed each on each pagm A lisfl ofl potentiaflfl usequ eguations is Km pagg 7i A periodic table E mnhained DH page 9. Score /20 @ g1 g1 /20 E mflkfiwgqt @ 1. Answer the lfollowing guestions about the ozone molecule D3. {a} (g points} Draw the two Lewis resonance Structures for ozone [Be sure to show all] Nalencd electron pairs. (b) points) Redrawl one a your m structures in the spacg [bo— lojnandindioatethdformalhhargeoneachofthethreeoxygen fisotopesl (c) (5 points) W hat as the Lelectrorfl arrangement orn the central oxygen atom and what as the m bond @ngle irfl ozone according to the VSEPR theory .1 {Q Q points! m is HE electron arrangement on the central phos— phorous atom and What is the Q—P—Q bond angle in the related molecule phosphorous dioxide according to the VSEPR theory .1 2J Klensider the W related molecules2 dichloroethylend [C2H2C12) and dichloroethynd (02012]. {a} (g points! Draw the Lewia structure of aich oroetEXne (C2C12 ) m indicate the value of either KIl—fl I—fll floond angld fusing the SZSEPB theoryl @ @EMEMafldthdbendsbetmnthem m atoms in W C2C12, usingj the Em W @ Be sure to indicate M hxbridization 0n the barbon atomsJ fifl —fpresen. (E points Draw the Lemisi strnctnre fer each] thhe three @ichloroethyleneJ CgHgClj, molecules and indicate the value of either fill—Q I—E] Bona ang in Leach Eolecule. (d) (fipointsfllesenbeirflmerdsmebnndsbetmeerflthdtmdharbhn atemsl fl dichloroethylenei C2H201j, usmg thd Valencd [Bend the: E: Be snre t9 indicate m hybridization (III the earbnrfl athnsJ ifl present. 3J flmsififl the neutrafl moleculart tfragmentj nitrogenl monohydrideJ NHJ {a} Q [Eointsy Md E1173 valence Bona wavefunction, w, for HE NH] Bonal Bd sure td incluae E1173 e ectron spinsl (b) (5 points] [Dravvt the Li IAQ tnolecular orbital energy flevefl diagram EtthHthlegnle mmmmfihydrogenflmw and anltrogen BE {[45 eV. Be m indicate thd m bflthd atomic m m m (c) (5 points) yom expect thd NH] bend tr) be olarize [2n IIIE polarized? Givd a cenerse rm m [based an thd mm tEeory (not electronegativity). (d) (5 points] [D0 Brow exfiecfl this fragment E bd fliamagnetid or para— magnetic? Cive a @1an Leasnrr E your ansmen EU Caiwhaje the bond orden from the MEAD—Mi) itheorfl for eacifl ofl thd following to decide iii the addition of an electrom E each bfl the following molegnles makes the mm more on less sjahle jihad the molegnle E [fiomtfi each] palri 2 for each bond orderJ El for comparison} E} mammmmmm (h) Nj is more or [le—ssJ EtaBle than N2— (C) m is more on less stable thaifl Bi 2— (d) Q is more on less stable thaifl UF— 5J [The ethylene @olecule2 [22H42 has a double bond witlfl a m bond lengthl bfl bpproximatelfl L31) m Estimate the energy of thd photog mmmmimmmjmmmmmfl Potentially Useful] Constantd & Eguations .. fl0/23/06 HZWEMEmolew 133 = R[NA = H.38066X10’23J FK’l E 2 @626 X 10—31L1 seo NA = 6.022! m M mole—l proton = M67263 M ED”? Q B 2 8.314 1 m mole—l fl. :1ng {[th 104713 0 2 H2197}! X] m m bee—l 6d : 8.8542 X 10—12 Coulomba E mfl elentron = 9.fl0944 X 10’31 1g neutron mass = 1 X] 10—2 [kig e ectron Cfiarge = 160218 X 10—19 £19111 6 = 56.705 nl/VWL’ZI{’4 em = eoh2/e27rrne = 5292 pm Fcoulomb= -m E= — dWK) E2112 A220 P/Al = <:1T4 {pm 2 W5 M=919293Mm60r Q—amfi: E(n)=7ZT8X10’18U;13l @(n7Z):—:ZE? AlElAlflZh[47T AlXApfihMfl ,h2 d2 mfig®+vflxlgfil=m§§l H(X) E M(X]@(X)dx \l'freeKX} = A] Emi27deeB fMWWfI/‘Wi = 72 gzmmm RH} = 109lZ3]cm‘1l p2 2m [71% J gm 2 @ bin(n7rx/L] E(nm) = Mlfi/SmL? VWQfl) = W Vt 2m 11/2Mfl — Ze2/47T60n H fl n = minim/xxx ct>| m = hm 1M = RH (mam — 1min) pzrny z2 [flaw] = l-—V13.60e fi g (njzeon2h2 [q1q2e27rm ZeffizZ—m Hglwuvm = Equ m = [13(nmnyan) = (n: : : n3> 2:3 ROW:le fl AElP‘“? : NA q1q2e2fl(47r€d d) All I A NA ((1101262/K47T60 K1 — €- m = (FIP::[EAJQ M=T6|Mflml arm—ml? MAJ—m = (Lfl02x/N N = EDEgAJQ— EBDEng) m = (#Valencd 6—D — [#LonePaM 6—D — g (#Bonding 6—] menE 04 3» mvEcnspuA t :88 $an $me :53 Among :mmv A88 Emmy E E gm mm 05 E u, # mm .E m: H: 0: m3 2: k: 9: E: «2 ma E ANNE mdfi «.2: 5pm: 8&2 am we mm in am mo 8 an E. mm mm 3.33 mod: 8a.: 3. H S 33 $8 25.3 a? m “8.3 no.5 Exam mom Em mi Um Sm dm PH. 02 £2 .HN .w hm nm E on t. 3 mv ww 5+ 3. 1‘ of an mm on [8% 3%. 3.3 :3 98.3 Sig main 2:: m «3.8 8.5 $3.. 8.2. mag LM mo 50 Hz 00 m.”— CE .5 b E um MD E mm mm mu mm .8 mm mm wm mm mm a an 3 2.38 $13 25.3 mg .8 fl min 8.3m E 6 a E 2 2 S m w n w m w m m2 .2 E E .1 E E 2 33m 33: :93 :3: $36 mama m2 ,m D m mm 5 3 m w m w m 38... 284 m E m. H w H ma wfi w d N m m H m H H ...
View Full Document

Page1 / 9

cem181f06e2 - Chemistry 181E Film @006 EXAM] II I...

This preview shows document pages 1 - 9. Sign up to view the full document.

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