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Intro Chemistry 1101

by: Alexis Tate

Intro Chemistry 1101 13699

Alexis Tate
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About this Document

These notes are just a recap of chapter 7 and help clear up any misunderstandings.
Intro Chemistry I
Alexander Schwab
Class Notes
General Chemistry




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This 8 page Class Notes was uploaded by Alexis Tate on Sunday September 25, 2016. The Class Notes belongs to 13699 at Appalachian State University taught by Alexander Schwab in Fall 2016. Since its upload, it has received 81 views. For similar materials see Intro Chemistry I in Chemistry at Appalachian State University.


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Date Created: 09/25/16
Ch. 7The QuantumMechanicalAtom Saturday,September 24, 2016  AtomicStructure - atomsare surroundedbyacloudof electrons(e ).Containanucleusconsisting of protonsandneutrons(p andn ) o What isLight? - Visiblelightisoneformofelectromagneticradiation ○Energy can betransmittedbetweentwoobjectsintheformofradiation ○Radiationhasan electricchargeand magneticcomponent ○Light travelsasa wave  λ (lambda)=wavelength(measuredinmetersusually) □ Redlight: λ = 620 -750 nm □ Greenlight:λ =500 nm - 550 nm ○Amplitude  Heightofthewave  Intensityoflight ○Wave travelsata constantspeed  c (speedoflight)=3.00x 108 m/s ○Frequency  Frequency=v  Numberoftimesawavetravelsupanddownin onesecond  Units:Hertz (Hz)or1/s (they'reequaltoeachotherHz = 1/s) - VisibleSpectrum ○"WhiteLight"containsallof thecolorsat the sametime ○Objectsreflectthecolorthattheyappear ○"Black" allcolorsare absorbed  c =λ(v)  Speedoflight=wavelengthxfrequency  Wavelengthandfrequencyareinverselyproportional(asonegoesuptheothergoes down) o ExampleProblem:  Thewavelengthofasatellitesignalis656nm. What isits frequency? c =(λ)v 3.00 x 10 m/s= 656nm x v -9 656 nm= 656 x10 m 3.00 x 10 m/s 656 x10 m9 14 v = 4.57 x 10 Hz  PhotoelectricEffect - o Metalsurfacescan absorb light(radiantenergy)andreleasee s(kineticenergy) o E is onlyreleasediflighthasashort enoughwavelength  Frequencyv.Energy o Metalseemsincapableof"savingup"Energy o In 1900, Max Planckproposedthatelectromagneticradiationwas"quantized" o In 1905, Einsteinproposedthatlightwascomposedofparticlescalledphotons  Natureoflight o Lightalso behavesasastreamofphotons  Photonsarebundlesofenergythatbehavelikeparticles  Each photonhas theenergy  E= h(v)  Energyof aphoton= Planck'sconstantxfrequency  h = Planck'sconstant(6.626 x 10 -34J s)(J=joules) o ExampleProblem:  A particlegivesoffanenergyatafrequencyof3.45 x10 Hz. How muchenergydidit giveoff? o E= h(v) o E= (6.626x 10 J(s))(3.45x10 Hz)2 -21 o E= 2.29 x 10 J o (TheHz (1/s)cancelsout withsecondsinPlanck'sconstant) o (Planck'sconstanthasunlimitedsignificantfigures(sigfigs)sotheanswer shouldhave3sigfigs)  Atomsemitphotonsw/specificenergies o Affectedbyatomicstructure  Wavelengthsoflinesinanyparticlewithenergycanbecalculatedby: o 1 = RH ( 1 - 1 ) λ ( n 2 n ) 1 2  n1and n 2re positiveintegers(thisisbecausetheyareenergylevelsofanatomand youcan't haveanegativeenergylevel) -1  RH= Rydberg'sconstant(109678 cm )  TheBohrModel o NielsBohr(1913)  Postulatedthate s travelindiscreteorbitsaroundthenucleus -  e s havefixedenergy(E2inaparticular-18it 2 E= (-hcR H(1/n ) =(-2.18x 10 J)(1/n )  e s can changeorbitsby absorbingoremittingphotons(energy)  LightAbsorption o Whenhydrogen'se s has n =1 (theelectronisatthe firstenergylevel),theHatom isin its groundstate Hydrogen Atom  e s absorbenergyEfromaphoton o n (energylevel)ofe goesup o ∆ E (changeinenergy)ofe =energyofphotonabsorbed o Hatom is nowinexcitedstate -  e has n =2  What isthe Eof thephotonabsorbedduringthistransition? -18 -18 Photon'sE= l1.94 x10 J l =1.94 x 10 J * ThisanswershouldALWAYSbepositivebecauseitdoesnothavedirection  What isthe wavelengthoflightcomposedofthisphoton? E= (h)v c = (λ)v E/h =v λ =c/v -18 15 8 -7 1.94 x10 J = 2.92 x 10 Hz or1/s 3.00 x 10 m/s =1.02 x 10 m 6.626 x 10 J(s) 2.92 x 10 1/s o Bohrmodelonlyworkedforhydrogen o Importantstepsto quantummechanics  Wave- particleduality - o e areparticles,buttheyalsobehavelikewaves(light)attimes  Diffraction  Electroninan Atom - o Positivelychargednucleustrapse s inatoms o Each e wavehasa singlecorrespondingE o Each waveis calledanorbital o Orbitalsareenumeratedusing3integers  n - principalquantumnumber  l -secondary(azimuthal)quantumnumber  m l magneticquantumnumber o n (principalquantumnumber)  Anypositiveinteger  Allorbitalswiththesamevalueofnare inthesameshell -  In Hatom, Eof e dependsonlyonn  In biggeratoms,theEof e dependson and m l o l (secondaryquantumnumber)  Dividesashellintosubshells  Valuesof are limitedbythevalueofn l  is an integerbetween0andn - 1  Thismeans can NEVERequaln  lis alsoalwaysa positiveinteger  Valuesof are denotedbyalettercode Valueofl 0 1 2 3 4 5 Letter s p d f g h  Subshellsarenamedusingnand ( i.e.1s, 2s, etc.) o Subshells  The1s subshellhasquantumnumbersn=1and l =0 o m (lagneticquantumnumber)  m lividessubshellsintoindividualorbits  Valuesarelimitedby l  m is an integerbetween- and + l l  So,whenn = 3, l =2, m =-l, -1, 0, 1, 2  Thisis a3d subshellsoithas fiveorbitals o Electronshaveaspin  e s havemagneticpoles  Spinisquantized"up"or"down"  Spinmagneticquantumnumberism s  m s+1/2 or -1/2  ElectronConfiguration o PauliExclusion Principle-notwoe s can have identicalvaluesforall 4 quantumnumbers - o Two e sin thesameorbitalmusthaveoppositespins  ExampleProblem: - o What isthe e configurationforLi? 1s , 2s1 * Electronsalwayshave oppositespinsinan orbitalindicatedbythe up and downarrows - o Hund'sRule - e s are spacedout as much as possiblebeforetheybegintopair up  ExampleProblem:  What isthe e configurationforNa? 1s , 2s , 2p , 3s1 *Forshorthandalwaysstart with thenoblegasright abovethe Shorthand:[Ne]3s 1 atom's row.Thisappliestonoble gasesso DONOTsay [Ne],say 2 6 - [He]2s , 2p  How manyunpairede s? *ALWAYS writeallofthe"up" arrows firstbeforeputtingin any "down"arrows.Sofor 1s 2s 2p 3s 1 unpairede - example,in2pthereshouldbe 3 "up"arrows beforeyoufillin 3 "down"arrows. - o Whene sare unpaired,thewholeatomactsas amagnet  Attractedto amagnet  Paramagnetic o - - Whene sare paired,e smagneticeffectscancelout  Notattracted to amagnet(repels)  Diamagnetic  ExampleProblem: -  Writethefullandcondensede configurationforPb. Pb:1s , 2s , 2p , 3s , 3p , 4s , 3d , 4p , 5s , 4d , 5p , 6s , 4f , 5d , 6p 2 2 14 10 2 Pb:[Xe]6s , 4f , 5d , 6p - 1 7  Whichelementhasthee configuration[Kr]5s ,4d Ru - (anomalyconfiguration)  Findingthee - o - Ψ(wavefunction)tellstheprobabilityoffindingthee someplace  Use dotdensitydiagramstopredictchancesoffinding e inaspot - - - o Aufbau Principle- e s fillthelowestenergylevelfirst;e s are usuallyin predictedspot 95% of thetime  SizesofAtoms o An e isattractedto protonsinthe nuclei  Forceof attraction(coulombforce:C)  Increaseswithincreasingnuclearcharge  Decreaseswithincreasingdistance o e s repeloneanother - o Difficulttoexactlyaccountforallrepulsionsandattractionsandpredictsizeof e cloud  Valencev.Coree s - o 1 Sodium:[Ne]3s o Valencee sliveintheoutermostshell  Calledthevalenceshell o Coree s are intheinnershells  EffectiveNuclearcharge - o Methodto estimatetheaverageenvironmentforvalencee s o Nucleusisshieldedbycoree s - o "effective"chargeofnucleusislessthanatomicnumber o Zeffuclearcharge) *Zeffjustthe attraction  EstimatingZ eff of electronstothenucleus o ZeffZ - S (protons)inanatom. Z is eff  Z = atomicnumber aFORCE  S = screeningconstant -  Usuallyclosetothenumberof coree s inan atom  Z effectsmanyproperties… o Z increasesfromleftto right onthe periodictableandincreasesasyougo up a column eff o As Zeffcreases:  Atomicradiidecreases…(sortof)  Meltingandboilingpointsincrease  Electronegativityincreases  Ionizationenergyincreases  TrendsinAtomicRadii o Atomicradiidecreasesgoingup ina givengroup  Numberofe s increasesasyou go downacolumn o Atomicradiidecreasesas you golefttoright in agivenrow  Zeffcreasescausingsmallere cloud -  IonicRadii o In general,…  An anionwillbelargerthanitsparentatom - -  Addinge s to an atomincreasese repulsion(makesitbigger)  A cationwillbesmallerthanitsparentatom - -  Removinge sfrom an atomdecreasese repulsion(makesitsmaller)  ExampleProblem:  Findthelargestand smallestatomorionineach of thefollowingsets:  Ge, Te,Se,Sn Smallest:Se Largest:Sn  C, F, Br, Ga Smallest:F Largest:Ga  Cr, Cr , Cr3+ Smallest:Cr 3+ Largest: Cr 2- 2- 2-  O, O ,S, S Smallest:O Largest: S  IonizationEnergy o Theminimumenergyrequiredtoremoveane froman atom - o Atomscan haveseverale s,and eachhas its ownionizationenergy(IE) o I - firstIE 1  Energyrequiredtoremovethefirste - *IEis justdescribinghow  Na-->Na + e - much ENERGYit would o I2- secondIE taketo removeane from -  Energyrequiredtoremovetheseconde - + 2+ - an atom  Na -->Na + e  IEforlightelements o In general,I 1I <2 … 3 o Coree s are verydifficulttoremove  1st IE trends o I1increasesfromlefttoright, bottomtotop -  LargerZ meffsit harderto removee s  Smalleratomalsomakesitharderto removee s -  FullyfilledsubshellsarestablesoraisesIE  1/2 filledsubshellsarestablesoalsoraisesIE  ElectronAffinities o Energychangeassociatedwithaddinge s togaseousatoms -  If∆ E> 0, thenthe atomis unstableuponaddingane  If∆ E< 0, thenthe atomis stableormore stableuponaddingane - o Roughtrend:increasesfrom lefttoright, bottomtotop QuickRecap of QuantumMechanics: (Dueto copyrightissuesIcan't includeapictureoftheperiodictable,sobearwithmeas I usethis sketch) (Rough)PeriodicTable Black arrows (increases):IE,Z , effctronAffinity Red arrows (decreases):atomicsize (Rough)PeriodicTable *Each rowhas a valueofn (Hydrogenwouldstartat n =1) and goesfrom1 to 7 l = 0 (s orbitals) l = 2 (d orbitals) l = 1 (p orbitals) l = 3 (forbitals) Thisis wherethealkaliandalkaliearthmetalsare(Groups1Aand 2A) Thisis wherethenonmetalsandmetalloidsare(Groups3A through 8A) Thisis roughlywherethetransitionmetalsare(Keepinmindyoudon’t - 2 start namingtheminthee configurationuntilafter4s and thenitjumps downto 3d) Thesearethe bottomtworows(lanthanidesandactinides)ofthe - periodictable(Keepinmindyoudon’tstartnamingtheminthe e configurationuntilafter…6s ,5d , andthenit jumpsdownto 4f)


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