Quiz_2_section_1B_answer

Quiz_2_section_1B_an - Name:AnswerKey Quiz2 October18,2010 TA:Robert 20ADisc1B 1 (theremaybemore thanoneanswer(10pts a n=3,l=2,m=1,ms= b

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Unformatted text preview: Name:AnswerKey Quiz2 October18,2010 TA:Robert 20ADisc1B 1. CircletheletterofthesetofquantumnumbersthatisNOTallowed(theremaybemore thanoneanswer): (10pts) a. n=3,l=2,m=1,ms= b. n=4,l=2,m=1,ms= c. n=2,l=1,m=2,ms= notallowed:m=l...0...+l d. n=1,l=1,m=1,ms= notallowed:l=n1.....0 2. Howmanyradialnodesarepresentforn=3,l=1?Howmanyangularnodes? (10pts) a. RadialNodes:nl1=311=1radialnodes b. AngularNodes:l=1angularnodes 3. Theeffectivechargesfornitrogenare:Zeff(1s)=6.66,Zeff(2s)=3.85,andZeff(2p)=3.83. Calculatetheaveragedistanceofanelectronfromthenucleusinthe2sorbital. (10pts) 2 1 ( + 1) n a0 r n = 1+ 1- Z eff (n) 2 n 2 2 2 (5.29 *10-11 m) 1 0(0 + 1) -11 r n = 1+ 1- = 8.24 *10 m 2 3.85 2 2 4. Theeffectivechargesforcarbonare:Zeff(1s)=5.67,Zeff(2s)=3.22,andZeff(2p)=3.14. Estimatetheenergylevelforthe2porbitalofcarbon. (10pts) 2 [Z (n)] n = - eff 2 (2.18 *10-18 J) n [3.14]2 n = - (2.18 *10-18 J) = -5.37 *10-18 J 22 5. Foreachpair,whichelementhasthelargeratomicradius: (10pts) a. b. NaandCl c. ClandCl1 NaandNa+1 6. Givetheelectronconfigurationforthefollowingelements.Pleaseusenoblegas abbreviationsifpossible. (10pts) 23p5 a. Cl=[Ne]3s 7. Beloware2randomnuclei.Drawthe1uand1g*molecularorbitals. (10pts) b. Mn2+=[Ar]3d5 8. FillinthecorrelationdiagramforH2+1,ahomonucleardiatomiccationicmolecule.Label theatomicorbitalsandthemolecularorbitalswithcorrectnotation(eg.1u). (30pts) a. Whatisthebondorderofthismolecule? BO=(BondingAntibonding)=(10)= b. Basedonthecorrelationdiagram,doesH2+1haveahigherorlowerionization energythanthemonatomichydrogenatom? Higher c. Basedonthecorrelationdiagram,doesH2+1cationhavealongerorshorterbond thantheH2molecule? Longer d. Basedonthecorrelationdiagram,doestheH+1cationhaveastrongerorweaker bondthantheH2molecule? Weaker 1=1010m 1m=3.28ft 1mile=5280ft me=9.109x1031kg mp=1.673x1027kg mn=1.675x1027kg CheatSheetPage rn=(5.29x1011m)n2/Z En=(2.18x1018J)Z2/n2 [Z eff (n)]2 n = - (2.18 *10-18 J) 2 n h=6.626x1034Js 0=8.854x1012C2J1m1 c=2.998x108ms1 e=1.602x1019C F=ma J=Nm N=kgms2 r n = n 2 a0 1 ( + 1) 1+ 1- Z eff (n) 2 n 2 BO=(bondingantibonding) MetricPrefixes 109=G 106=M 103=k 101=d 102=c 103=m 106= 109=n 1012=p 1015=f F(r)=q1q2/40r2 V(r)=q1q2/40r=Ze2/40r E=h=hc/ =h/mv=h/p ...
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This note was uploaded on 01/31/2012 for the course CHEM 1 taught by Professor Liu during the Spring '11 term at UC Davis.

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