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General Chemistry

by: Austen Pollich

General Chemistry CHM 11100

Marketplace > Purdue University > Chemistry > CHM 11100 > General Chemistry
Austen Pollich
GPA 3.75


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This 24 page Class Notes was uploaded by Austen Pollich on Saturday September 19, 2015. The Class Notes belongs to CHM 11100 at Purdue University taught by Staff in Fall. Since its upload, it has received 34 views. For similar materials see /class/207977/chm-11100-purdue-university in Chemistry at Purdue University.


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Date Created: 09/19/15
Hour Exam 1 Study Guide Ch 1 Matter and Energy 1 Matter and its Classification Need to J39 quot 39 39 homo vs 39 39 vs elements I pure vs mixture Pure Substances Have uniform or the same chemical composition throughout and from sample to sample Two kinds of pure substances Elements An element is a substance that cannot be broken down into simpler substances even by a chemical reaction Elements are separated further into metals and nonmetals Compounds A compound is a substance composed of two or more Compound a substance composed of two or more elements combined in definite proportionsaka law of definite proportion Molecule is composed of atoms elements combined in definite proportions 2 Pthical and Chemical Changes and Properties three states of matter gsolid liguid and gas Useful physical properties mass volume density and temperature their units How to calculate density amp y lume And how to convert among temperature scales Table 12 States of Matter ijla Saoliril Liquid Ga 7 3 5 32mg hcc rolnx lame r quot lmy cin a i39 ZS rv 514 Ed 21 3 p559 r iVvjm 3 y lrrmh 1 r Volume amount of space occupied 7 Volume length Width height 7common units cm3Emilliliter mL iLiters L Density ithe ratio of the mass to its volume density mass V0 ume iUnits gmL solids and liquids or gL gases Temperature 7a measure afhaw hat or cold something is relative to some standard imeasured with a thermometer 7 at which a phase change occurs is independent of sample size funits are degrees Celsius OC degrees Kelvin K and Fahrenheit oF TK T C 27315 T F 18T C 32 3 Energv and Energv Changes kinetic amp potential energy we Energy capacity to do work or to transfer heat Kinetic energy the energy of motion mechanical Potential energy energy possessed by an object because of its position chemical electrical etc 4 Scienti c Inquirvl quot quot 39 39 Observations include experimentation collection of data A hypothesisis a tentative explanation for the properties or behavior of matter that accounts for a set of observations and can be tested A scientific lawdescribes the way nature operates under a specified set of conditions Theoriesexplain why observations hypotheses or laws apply under many different circumstances Ch 2 Atoms ions and the periodic table 1 Dalton s Atomic Theory four 39 1 All matter is composedof exceedingly small indivisible particles called atoms 2 All atomsof a given element are identical both in massQand in chemical properties However atoms of different elements have different masses and different chemical properties 3 Atoms cannot be created or destroyedin chemical reactions 4 Atoms combine in simple fixed wholenumber ratios to form compounds 2 Structure of the Atom memorize de nition of Z and N 39 properties of protons neutrons and electrons I 1 among observation and theorv memori e Thompson and Rutherford e oeriments 7 know the 39 39 but not experimental details Protonp Q 16022 x 1019C expressed as 1 m 16726 x 1024g hydrogen atom 39 Neutronn no charge m 16749 x 1024g Electron e Q 16022 x 1019C expressed as 1 m 91094 x1028g 11836 of a H atom Atomic NumberZ number of protons identical to number of electrons Neutron Number N number of neutrons Mass Number A number of protons and neutrons Isotope of an element an atom that contains a specific number of neutrons an element may have multiple isotopes Isotope symbol 3 Ions formation of cation and anion39 charge of Ion forms when an atom gains or loses electrons 6aining electron more e than p yields anion negatively charged Losing electron yields less e than p cation positively charged 4 Atomic Mass de nition Isotope de nition Symbol Nature Abundance Relative Atomic Mass how to calculate atomic mass unit amu Mass of 12C 12000 atomic mass unit 1amu 112X mass of 1 12C atom 16606 x 1024g Individual atomic masses are determined by mass spectrometry mass numbers on the periodic table are relative atomic masses Relative atomic mass is the average mass of the individual isotopes taking into account the natural abundance xiof each relative atomic mass sum the weighted mass contributions from each isotope A x1A1 x2A2 x3A3 xi is the natural abundance of ithisotope in decimal a ions A is the exact mass of ithisotope in amu Relative Marnie Mass Examples 155 rugt Mass ljanvui 431 Hn rritnuhun HI Ag L Eli905 39 1532 39 39 My 10333133943 34615 F xlltt 2 E in 7quot 1 5 Elemen39l li39h39Lr s tampond m Inn usnl upu L was l39i l ubu dmze 0137 3 am I I39thmi39 imusrs 7331503 atwc qrcc 0931 jglw al a in re lnl ire ulnrnh rms39s 2quot 5 The Periodic Table main gzoup transition metal39 metal metalloid and nonmetal The elements are arranged by increasing atomic numberZ and in columns and rows to emphasize periodic properties The columns aka 6roupsfamilies are designated in two ways A Roman numeral I through VIII and a letter A or B OUR Choice or An Arabic number 118 The rows are called Periodsand designated by an Arabic number 1 7 Main5mm Elements trivial names 51 i u layv E Ha39vzn a i Group I or II tend to lose 1 or 2 electrons respectively Group V VI or VII tend to gain 3 2 or 1 electrons respectively excluding metal loids metal 6 Math toolbox 11 pp 35 36 7 how write numbers in scienti c notation39 simple calculations w scienti c notation Ch 3 Chemical Compounds 1 Ionic and 39 39 and Nonelectrol es ElecTrolyTes subsTances ThaT release ions when dissolved in waTer This process is also called dissociaTion or ionizaTion conducT elecTriciTy NonelecTrolyTes subsTances ThaT do NOT dissociaTe in waTer 2 39 and I 39 Ch02 Exam 2 Study Guide Chapter 3 Chemical Compounds 1 Ionic and Molecular Compounds covered in T1 Table 31 Ionic vs Molecular 1 pthical nronerties39 Electrolvtes Ions charge of ions from Imic mpnunds IquotIIllinr n 39l lire gr HI 5 1 Urn hujh 3915 hm pain 39va high t r my ravinl Molecular Cornpr 33 HHILIKJ r salic39 50f inliv lam mzll a 973 ng auxinr palrl oN cars 51 k i39r ryrrcl aim 39u r i at an l39lzzllr39uu 4uuu iv is you Waivqu OH LLI n 3 J era m vr pnnrl ti 39w rm Iquot pun auu m ulquoti l m 39i quotu i39rr m u39l v3939 r azr 2 Monatomic and Polyatomic Ions a Focus on monatomic cations anions b For polyatomic anions see Ren s Rendition 8th lecture 7 memorize those in red 3 Formulas for amp Naming Ionic Compounds Charge neutrality for correct formulae name cation first then anion Roman numeral for variable charge on metal 39 Simple caTion same as The elemenT eg Na sodium Ca2 calcium Cs cesium 39 Simple anion use idequot suffix Cl chloride 52 sulfide N3 niTride PolyaTomic ions oxoanionsAOnm Remember aTequot eg ClO3 is chloraTe One less 0 aTequot becomes iTequot eg chloriTe is Cl02 One less 0 iTequot becomes hypoquot eg hypochloriTe is C lO One more 0 aTequot becomes perquot eg perchloraTeis C lO4 CaTionfirsT Then anion eg NaCI LiNO3 K20 Li2503 Formula charge neuTraliTy ToTal posiTive charge ToTal negaTive charge O Eg Formula forliThium niTride sodium oxide ammonium sulfaTe calcium phosphaTe magnesium hydroxide Use Roman numeric To noTe The charge if poTenTially ambguous Eg Fe2 ironII Cu2 copper II Cu copperI Old naming sysTem exisTs for TransiTion meTals and Sn and Pb based on The rooTs of The original LaTin names The smaller charge has an ousquot ending The larger charge has an icquot ending Tnhlr 3 tr Namn nr Tani Commind anTnininj Maral uiTh 39JJrirjlglg Elna qt nerFauan Inn SYHQFIGHE Nnmr Trivial Namr Fat Fe 393 rnnlfLEII chiavid hummus mlw udt Fail 4 inronfEEI gland i39lcrirh chloride Euz Cuquot Lappcrw l and pgau mm M Euquot tuppw l mud cupriv mid 5m Em l 1191111 axial wrannnus oxide ERIE I 5mquot tin h mind I si unni mail 4 Naming and Writing Formulas for 39 39 Binary Compound conTains aToms or ions of only Two elemenTs Naming molecular compoundsamp 1 Name The lemeosT elemenTas we would a main group meTal according To The periodic Table 2 Name The righTmosT elemenTas we would a monaTomic anion drop The ending of The name from The periodic Table and add an idequot 3 Wmmmnhmmmmm smgywcmm rul2wphls a m ap w my mwmm mm 4 Us em Wm n m m mm M mamsacmmmm mu 3 a cumquot m Pmlvxu mm Nam mm NW 1 ha a m r 2 mm 7 7 a m a 4 nn 4m 7 n mm s marmsm mum um 62mm hnnumsdu m uh mamanmAWMsm mmmnOBmgmA mm b ukwmxfmm uphz17hbk z y a as and as 5 subs uncls hu mmmm m Name w m mm m tame and w mm mom Mm m1 Mamw a I 1 mm m n lln39hlnm 39Ac ds m nund mummy n m um WWW m mm us hydraJaHauId by m m hllkmm Endlhmm mm x an M and 39Ac ds WWW WWW mm m mm by mm m m a m WWW mm mm upmwwm m upmwwm 7 mm mm m WM and m m Ind quotwinn lo h I Cquot 39 i C39 U Lquotulumtt CIDy patch aquot I39 C Chapter 4 Chemical Composition 1 Mole Quantities 39 Avogadro s number what does it represent Contains 6022 x 1023particles molecules atoms ions formula units etc This number is called Avogadro39s number The amount of substance that contains as many basic particles as there are atoms in exactly 12 g of 12C 2 Moles mol Masses and Particles molar mass gmol Describes the mass of 1 mole of a substance Obtain the Molar Mass from the periodic table by assigning different units to the atomic mass Molar mass is the conversion factor between mass and moles mol mass molar mass 39 mass mol x molar mass 39 molar mass mass mol Graphically the relationships are shown below grams of substance moles of substance numbers of atoms molecules x molar mass divide by molar mass x 6022 x1023 divide by 6022 x1023 Applications Scenario 1 given mass determining moles divide mass in g by molar mass HW7 Q8 Q1 lb Scenario 2 given moles calculating mass multiply moles by molar mass HW7 Q9 Q1 1a Scenario 3 given mass calculating number of molecules step 1 divide mass in g by molar mass to get moles step 2 multiply moles by Avogadro s number HWS Q3b Q3c Scenario 4 given mass calculating number of atoms of a specified element E step 1 divide mass in g by molar mass to get moles step 2 multiply moles by Avogadro s number step 3 multiply by the number ofE in the formula unit HW7 Q10 Q1 1c Scenario 5 calculating the mass of a single atom or molecule in grams step 1 molar mass of the atom or molecule step 2 divide molar mass by Avogadro s number HWS Qla Q2a39 T2 2008 Q8 amp Q15 Scenario 6 given number of molecules calculating the moles and mass step 1 getting moles by dividing number of molecules by Avogadro s number step 2 multiply moles by molar mass to get mass in g HWS Q1b Q1c Q2b Q2c Q3a39 T2 2008 Q9 3 Determinin anirical and 39 39 Formulas Empirical and Molecular Formulas Molecular formula in the actual number of atoms Empirical formula in the simplest ratios of atoms in a compound Empirical Formula from mass Assume having 100 g 70 of element g of element Convert from mass to mol Find whole number subscripts Molecular Formula from Empirical Compare exact molecular masswith empirical molecular mass mass calculations from either given masses or formula mass empirical formula molecular formula with molar mass info Percent composition of C6H1206 glucose C 400 H67 O 533 4 Chemical C quot39 of inntinn 39molaritV M Solution homogeneous mixture at molecular level made of solutes and a solvent Solu re present in a lesser amount Solvent the substance that dissolves amp in large quantity Concentration amount of solute relative to solvent Percent by Moss Expresses concentration via percentage 70 mass mass solute x 100 mass solution Molari ryN number of moles of solute per liter of solution M moI V DilutionThe process of adding more solvent to solution M mol V mol M X V V molM watch out for the unit of V L Versus mL Law of dilution M X Vconcentrate M X Vdi1ute Also covered Math Tool BOXes 12 amp 13 7 pay attention to relevant homework questions and Q1 Q7 from T2 2008 Chapter 5 Chemical Reaction and Equations See my week08 post Exam 3 Study Guide CHAPTER 5 55 Predicting Chemical Reactions 1 Know how to classify types of chemical reactions Decomposition Combination SingleDisplacement DoubleDisplacement 2 Know how to write a balanced equation when reactants are qiven Class at chemtoat Reacuons mass mum mums Emu mmmtun on s cm Combmmmn u gt g At All sngu tarmm mum AACD cum DtsmammLnl 1 cnmpoum 1 compound Doumc Dtsbhnnmmxl Zen pounds Zcumpnunds co gtEr cr oz Decumpusmun a Decnmnasmnn at metal carbunale m msla mde m carbnn mam 55 Mun Moods 4 Nt0s m gt a ommpomn at mom Nymuxxdcs Io maul nxmns m wmw CJ39OHLLM CzOlst H Olm u Demmpusmnn at hydrate m anhy ruus umnnunds and walzr CUSOA EH CKst cuscnst smog u Dtcumpusmnn a nxtdzs m humus at Au Pl and Hg 1 WE ehzmmls PYCtht mm A 0le Cumbmatmn Dunng a Combmauon reacuon two substances combme to form a smgte compoun ltMost metats react wtn most nonmetats to form tomc compounds ltA nonmetat may react mm a more reactwe nonmetat to form a motecutar compo nd ltA compound and an etement may comome to form anotner compound tt one extsts wtn a mgher atom atom r t at o z anamvn 5025ng can E grednctwhchmelalsxeanwxmacxm An WW Athnananm 52112515 we can u dmwhzthznhz mug 1m 7 Sng Dwsp acement K Ba mm W Sr y E m m m H ymnm on T2 nanny mcreasmg a 5 unuemanu Wee yges m duume msg acemem veacuuns Pvemgnauun Reacuuns u Gas Furmauun Reacuuns m AcmEase Dan 2 Dwsg acememF vemgw a mm a Memunzemsummesans Immupr um unal csc5rsc Easo AqSD HqscPnso Hmmesmgx ng ng ltbX30lt urvu Sul de uxdes and hymmme 10me Xhosa a Na39 K NH aur Mum mm mm m somm sans Snmrnn aming a w NW Sahscwla quotgm and ammo GasFuvmauun m same doub erdwsplacemenl reactmns 2 gas 5 armed whmh hams dr a he reactinn In common cmEase An acid reams mu 3 base to form an omc compound 58 and wa er 2mm 7 ummammmeamngunumnummacuuns Canyuuwme ba anced eguauunsm cumnusnun reactmns Wnen a veactam 5 Wm oumnumn A cambus un reaction 5 a 5n type ar reaclmn that does nut a ways an mm one a he earner caxegones In a combushon rancher a subslance reacts th oxygen m a reaction that burns m produce heal am Name The mast cammnn cambuslmn reachcns mm the burning nfa nyamcarbam era hydrocarbon comame oxygen Thu prudms am d ways Larbun mama and water mum 5 5 Rnmmlna Rueuam In Aluuunl soluunn a Knuwme meamng ummecmayeguauuns mmc eguahuns netmmc eguauuns and 592mm ms Malecular egua m AgNO1aq Nammq a AgCKs NaNOaaq Innlc egual 7 WS soluble Iomc compnunds as separam Ions Aa39Iaq v NO aq v Na aq CHaq a AgCKs Na aq N03124 AgCs39 nsmume cumpmmds dc nm em as ms smce he spsclalonons do not bhange we can leave them cut loe nct mm EuUE on A9391aqcquotaq A AQCHS a When a veacuun m aqueuussmmm sgNen canyuutemech mns undevgu W I ill umlllnn 1 Knuwmemeamngma nameueguanun 1 53 alumu Euunlcn Halolulu common Human 39 The process at delermlnlng the amount uf a reactant or product lrom another reactant or product tn a reactron ts called storchtometry 2 yyhehthe guantly erthermol or a otohe leactam ulgmductlsgwen cahyou calculate the guantltles otthe lemalmng leactams aho gluductswhen a balanced What rsthe cohyersroh tactortor molemole comersohv A male ratto rs used to relate the number otmoles at one reactant or product to another What rsthe cohyersroh tactortor molemass cohyehsohh We convert grams to motes uslng molar mass Next we relate moles of Clz to moles of Na uslnq the mole rauo The last step ls to convert moles otCtz to grams usthg the molar mass 0 l2 Formassmasscomersrohhhat rsthethstthhoto em 3 yyhehthe uarltl otreactaht s ahothe e otreactroh ls ryeh e chantr u leltlny Rustm a When leactamsare nutmlxed tn a stotchtometnc manner canyou toenthythe ltmtttne leactanw ten two reactants are mtxedt one usually does not react completely because there ts too much at tt i The lrmtttng reactant ts the reactant that reacts completely and is thcretorc not present when the reacllon is complete Slnce the ltmrtrng reactant reacts completely lts amount eelermrnes the amountolproducl that can lcrm WWW amount ot gluducts7 can you also gledlct the yge ano amount otthe reactant that tematns unreacteev chapter 55 PIrnMYIIId The amounl ol producl aslually obla rneu ln lne lab aclual ylaldl ls usually less than lne amnunl predlcled by calculalrans lneorelrcal yrelu Vlelns desmbe lne amounl nl pmduct and can be m mass unit moles m number nl rnalacules Thqyyugm mg descrlbes how men afa pruduct ls aclually a alned relalrye la lne arnaunl lnalsnauld arm assumlng complala reacllon or he lrrnrlrng reaclanl P W W Acluamald OW me 3 Then llcaneld quot yrelunlgrnuuclnularneursgyen canynucalculalelnegercenlyreluv 1ns12p Delerrnrne lne lrrnrrne leader In slep aaseu unlne lrrnrrne leaclam calculalelnelnenrelrcalyrelu nlne nrnuucl 3n slep Usrnelne acluaneld anu calculaledweld calculalE percenlyrelu my small cnanllu a Knuwlhe energy change anu neallransrer bemeenme gnemanulne sulluundlngs rnynyeuwrn Exulhelmlc anu enunlnerrnrc leaclluns When ene y ls released lmm ha atoms or a cnarnlcal r sl erred lo me surmundlngs Tna amounl released by lne reacu an ls equal In me arnaunl absnrbed by me surraunurngs Exalllermlc Reacllon a reacllon lhal releases energy cnernlcal Energy at Reaelanle e Chemlcal Energy at Pro ucls EndalhermmR e areacllonlhzmbsnrbs energy cnernrcal Energy of Readaan lt chemreal Energy or Pronuels Knuwlhe meanlng all game neal Sgeci c heat The amount of heat required to increase the temperature oti gram et a substance by 1 quot C Specific heat units Jiig DC or camg DC 10 Know the equation used to calculate the amount of heat q What is the meaning ofthe sign of 9 Amount of heat lg released or absorbed by a substance q mCAT m mass C specific heat 3J2 temperature change if 2 it Ti 11 When a substance is immersed in water in a calorimeter can you calculate the amount of heat released or absorbed by the substance when the temgerature chanqe mass and speci c heat of water are known Calorimetry is used to determine the heat change ofa system by measuring the heat change of its surroundings in this calorimeter an insulated cup is used so the surroundings is limited to what is inside the cunt qsyeiem qsurreundings D 67 Heat Changes in Chemical Reactions 12 When the heat change per mol or dram ofa reactant is given can you calculate the heat change for any given amount of the reactant The heal change tor a chemica readlun can be determmed m the same way as The heat change for the p ECe ofcopper pre if the reaehon lakes mace in suman remen qsurmunumgs any Gum m Chute 7nd any x Ehm1 l ectmmggmm Radumm u mm xehnanshpbethenthz wmlzng hmdthz fnquzncy u mm xehnanshpbethenthz wmlzng h are fnquzncy mar phnmnemxgy lugm IS a form 0 energy cauea elecvnmagneuc radwauon m rad am ener ngm IS a form a energy caued electromagnellc ramauon arramanleneruy Wave ength v The mstanee between we currespundmg pumts an a Wave mt are meters rn ur ummun y nanumeters nm u a m Frequency U A rneasure enhe numberufvvave ydes thatrnuve thruugh a pumtm space M s Umts are henz Hz Whmh are the sarne as mverse secunds 15 Wave ength and frequency 5 mverse y prupumuna a v cV v e c speeumhghuanmnxrns 2 Understand Wm 3911 mime hmsmthz ammo sgctmmzan Enmmm hv cl EWM 7 energy a me phamn m James 1 anck s canslam 6525 x10 3 Is 1 used m Ugh 30 x1e ms I avelenglh rn m The hm speurum death lament Ts umeue so u can be used 0 idenmy Lhc e ement 29 ewemenhs ngerprint 3 Uhdzxstnndthz Ea mndzlafthzh meta Eteetmns arm the nueteus H1 spectltc ctrcular p2 ways hts means the etectmns can have only cerlam energtes Energtes are quanuzed he energy tncrcases as us mstance tram hn nucteus tnereases Bohr tahated me etectmn arb tls wtth a numbeh starlth wtth I ctesestm the nucteus and Increasth as the urb tls get runner away mm the nucleus When the etectmn maves from a highsrrenergy man to a Iewanenergy 0th tt teses energy equat lo the mflerence m energy between the amt E Em A Emmat The energy ts released as a pheton the emtsstcn a hght afsublzvelsm nmlg stlblzvels Esgblz faxeuhmmtglnnmhex nnmhexafm umlsm chm 1m maximum nnmhexalechmns meachat unaJsublml The modem medet at the alum ts based on Schmdtnger39s mathemaktcat deEI oi wa The energies of etackmns are SUII quanttzed but lhls madel descrtbes etectmhs as accupytng nth ttats net arb tls Orbtlats 1ree utmenstonat regtons in space where eleclmns are hkety m be found not a eucutar pathway Pnnclpal Energy Levels Orbllals ufslmllar SIZe exrsl ln e same prlnclpal energy level n s the pnncrpal energy level mcraases lhe armlal exlenrls further hem the nucleus There are four general types oiomnals s p a and f s ermlals Sphencal shape 9 urbllals Dvrnlmell shape Three 2 ermlals In sunlevel PPPyF They lle perpendreularle nne anelher These urbllals everlap wrlh lherr oenlers at lhe nucleus at urbllals Flve u ormlals ln 3 svnlevel furbl a s Seven l erbllals ln a sublevel The rsl pnncrpal energy level n1 s sublevel 1s ernrlal The second prlnclpal energy level quot2 sp sublevels 2s amnel lhree Zp ormlals The mlrd pnncrpal energy level n23 spd sublevels 3s orlmal lhree 3p orbnals hve 3d urbllals fmurlh pnnclpal energy level n24 spglsuhlevels 4s arbnal lhree 4p ormlals hve 4a urbllals seven Mammals 5 mehnwmdnwm uml mum elzcl mmccan moans Aurbau grlnclgle Electrons ml omltals starting wrth the lowestenergy urbllals Pauli excluslnn grlnclgle A maximum oflwu electrons can occupy each orbital and they must have opposlle s ms 5 rule Electrons are distributed lrlto orbllals of identical energy same sublevel In such a way a to axt her at unpaired eleclrons it takes a ll le bit o energy to pair up electrons so tl l single electrons occupy dlfferenl orbitals cl the same energy glwtlitate he arrangement of electrons in lhe lcwest ene slate at Excited state Notice lllal lite number cl columns n the s n n and lblccks is the same as the number nl eleclrcns aHuwed ln each subleyel The principal energy level numb n the number that comes belore lhe subleyel letter designation is the same as he peneo number lor the s and p subleyels The last lllled pnnclpal energy level is called the valence ievei or valence shell alellge electrons occupy orbitals in the valence level AH the other electrons are called core a eclmns ur Inner e eclmns Because the valence electrons are fanhest from the lhe nucleus they are lhe cnes lhat participate in chem a reactions Elements in the same group have the same number cl valence electrons ln ions the number of electrons does not equal the atomic number We must add or subtract electrons depending on whether lhe ion is an anion or cation u Mmcme an emxgyaxdzxafm nmls lstlt2plt3 3plt45lt3dlt4p u Can ym carmun nnmhexafnnpumdelzcrmns andwlence elzchmqs mm um dugrams Dr 212 m canl39 munng em y yarence electron lram a aseaus alum amzalmn energy mcmases up a group because the valence clemmns become rncreasrngly c oser to the nucleus Jonizauon energy mcreases mm new to new m 2 pence ecause me va ance erecxrons are mamasing y held more ugnuy by the nucleus which Is moreasing m me number nl prawns Generally IEA gt 152 gt IEy 2 mem catmmc mamam ndnls ls Menmfmmthz ammlc ndnls Caucns are smaller than unerr neutrar alums e g j K K Anrens are larger lhan mew neuual mums a g F lt P Fur nny saehsclmnlc genes as me numberur proluns rrrcreascs we ran srza decrmascs e g5139gt cw gt K gt Car Fcrrscelocrrcnrc caucnsas the charge mcreases he ronrc srze decreases Forisoelectmm c anions as the charge Increases becamas mare negauve Lhe mmc srze mcreases pm 1 Undzxstnndthz chandensnc feahlres afmmc andsvwlzmbands 0an cm Occurs between a metal and a nonmetal Electrons transferred from a metal to a non metal forming a metal cation and a non metal anion each with a stable noble gas electron configuration The ions are held together by electrostatic forces Each element immediately following a noble gas is a metal Metas lose electrons forming a positive charge to become cations Each element immediately preceding a noble gas is a nonmetal Nonmetals gain electrons forming a negative charge to become anions Formation of ions and ionic bonds relates to an element s electron con guration Many maingroup elements either lose or gain electrons to become isoelectronic with a noble as Covalent Bonds Occurs between two nonmetals A bond created by the sharing of electrons between atoms no electron transfer between atoms Electrons typically shared in pairs 2 Know the difference between the polar and nonpolar covalent bonds Nonpolar covalent bond Equal sharing of electrons Occurs only when the same kind of atoms are covalently bonded Polar covalent bond Unequal sharing of electrons Occurs when different elements are covalently bonded to one another Different elements have different electronegativities This result in unequal sharing which can be described as partial electron transfer Partial charges are developed on the atoms sharing electrons Polarity The degree of transfer of electrons in a covalently bonded molecule composed of different elements atoms 3 Know the tendech of Incre lL in the periodic table 4 Know how to use electrone ativit to determine the olarit of covalent bondin The greater the difference in electronegativity the greater the ionic character and the more polar the bond thatjoins the atoms No difference in electronegativity between atoms in a covalent bond results in a nonpolar covalent bond 5 Know how to draw the Lewis symbol TABLE 83 I Lewis Symbols of the Period 2 Elements Group Ll IIA2 IIIAUS IVAI I VAlS VIA16 VII17J VIDAUII Electron cunligmnlion IGZA Iflflp39 IIIJZAlz z I Valance elrclruns l 3 3 4 5 f1 7 3 Lewis symbol Ll39 39Be 6 Understand the Octet rule Just as in ionic bonding covalent bonds are formed so that each atom can have the noblegas electron con guration Covalently bonded atoms achieve 8 valence electrons by sharing electrons H atoms bond with other atoms to obtain a total of 2 electrons like He Sin le Bonds One pair of electrons are shared by two atoms 1 Molecules with an odd number of electrons e NO I Incomplete octets e or BF 7 ow the characters of ionic and covalent compounds Ionic Ions of like charge repel each other and opposite charged ions attract n39 39 quot 39 quot in a regular quot39 39 39 quot r crystal lattice that maximizes the attractive forces and minimize the repulsive forces TL L AsizesofAu39 at Au writ ionic crystals 1 crystal lattice valent o nonmetals form a bond the bond is covalent They are both close to the noblegas electron con guration so sharing will allow both to obtain i In a covalent bond each shared electron interacts simultaneously with two nuclei Covalent bonds exist only between the atoms in a molecule Molecules are not covalently bonded to one another Melting and boiling covalent compounds molecular compounds do not involve breaking covalent bonds1 Therefore these compounds have low melting and boiling points They o en exist as gases or liquid at room temperature 8 Know how to draw Lewis structures for covalent compounds 1 WRITEAN ATOMIC SIltELETON THE ARRANGEMENT OEATOMS ARE USUALLY SYMMETRICAL T CENTRAL ATOM TENDS TO BETHE ONE THAT IS LESS ELECTRONECATIYE IS RRESENT IN THE LEAST OUANTITY AND CAN EORM THE MOST BONDS SUM THE YALENCE ELECTRONS EROM EACH ATOM TO GET THETOTAL NUMBER OF YALENCE ELECTRONS 3 PLACE TWO ELECTRONS A SINGLE BOND BETWEEN EACH RAIR OE BONDED ATOMS 4 A REMAINING ELECTRONSTO COMRLETETHE OCTET OE EACH OUTER ATOM AND THEN TO THE CENTRAL ATOM IE NECESSARY AND IE THERE ARE ELECTRONS AYAILABLE IE NECESSARYTO SATISEY THE OCTET RULE SHIET UNSHARED ELECTRONS EROM NONBONDED ROSITIONS ON ATOMSWITH COMPLETED OCTETS TO ROSITIONS BETWEEN ATOMS TO MAIltE DOUBLE OR TRIRLE BONDS 9 Know the relationship between the number of valence electrons and the number of bonds that non metal elements can have 10 Know when double and triple bonds can form How many valence electrons does an oxygen atom have How many does it need to obtain an octet 02 has a double bond two pairs of shared electrons How many valence electrons does a nitrogen atom have How many does it need to obtain an octet N2 has a triple bond three pairs of shared electrons ll Know the definitions of classes of hydrocarbons shown in Figge 820 Compounds containing hydrogen and carbon Aliphatic hydrocarbons Linear molecules with single double and triple bonds Alkanes contain only carboncarbon single bonds Alkenes contain at least one carboncarbon double bond Alkynes contain at least one carboncarbon triple bond Aromatic hydrocarbons Molecules containing carbon atoms arranged in a sixatom ring with alternating single and double bonds resonance or delocalized structures All Hydrocarbons are Nonpolar You do not need to memorize Table 84 which we did not learn 12 Learn how to predict parent structure and molecular shape using the VSEPR theog VSEPR theory is based on the fact that negative charges repel one another Line drawing for bonds Solid lines In the plane Wedgesln front of the plane Dashed line Behind the plane Parent Structure atomse39 2 linear 180 atomse39 3 Trigonal Planar l20 atomse39 4 Tetrahedal 1095 Actual Shapes Linear atoms 2 unshared e39 pairs 0 Linear Trigonal Planar atoms 3 unshared e39 pairs 0 Trigonal Planar atoms 2 unshared e39 pairs l Bent Tetrahedral atoms 4 unshared e39 pairs 0 Tetrahedral atoms 3 unshared e39 pairs l Trigonal Pyramidal atoms 2 unshared e39 pairs 2 Bent Also memorize bond angles for each shape 13 Learn how to distinguish polar and nonpolar molecules Molecules with non polar bonds are non polar Molecules with polar bonds may be polar or non polar depending on their shapes or geometry They are non polar when polarities of polar bonds cancel out They are polar when polarities of polar bonds do NOT cancel out The polarity of the bonds and the polarity of the molecules may be different Polarities of polar bonds cancel out I This happens When a central atom has no unshared electrons and are surrounded by the same type of atoms in a symmetrical manner linear trigonal planar tetrahedral A rule of thumb for solubility is like dissolves like which refers to similarity in polarity Ionic salts and polar liquids dissolve better in polar liquids than in nonpolar liquids Nonpolar liquids dissolve better in other nonpolar liquids than in polar liquids D Know the difference between polar bonds and polar molecules D Know when the molecules containing polar bonds can be nonpolar


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