Notes after midterm 3 until the final
Notes after midterm 3 until the final Chemistry 1220
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Date Created: 06/04/15
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Ewing unm 4x 5m3r umuhn munquotynu an Hum E quotI 45quot uunnpu quota 93 3an 14 um namuwn JE n mamas 5 HT mu uanuqnwu a muma3 r Fu ib FE mun 3 v5 Fania 35 w 5a ILH qua n wmma 3 h n n T u n q n E r ma a rm 93me t M umu ha y ham313 rm w Ha dmh qu J warn auru Haw u 4 Hamqu 13 Jam av 11 u M wail Hathaw 1 Hum an 1 wwfha m Fu w mulm w a u an v1 a nu 5 qu nay HHS 1H HEWIUE Hrfam mud jam Edmu u w 5 5quot Fi atquot HE E a T W 5 EH u w 13 g 2 r1n 1 5 5533 12 3 an 7 a 5 is 3 i H 3512 Eu Um HE in 3 43 an n ES 3m mun Era I m HS 3 65 mm WEE mmva m um 33 Wi ui DEE mxma w mva linen 93 Emm lal lhu n thg r REEFEllquot firmIE Tmi39 l il 39li Eula435531335512 Endmiquot Ell mlm39 39J gm Ilia ijn 39rir EFli M ohmmm in um w m lll lliii f lh TM F39IEPiIIEL39u Him will I J JJEI FIEIla 315 IHI quotHi ll ll rr 3 i ll H5 W l i i m l l IlaIa iII IIu Hula Il airIla w Midterm 1 Midterm 2 Mhidterm 3 A 320 A 305 B289 B 272 B 255 C 238 C 218 C 198 D 178 D 158 REEL lab Lecture41015 Which complex ions isare most likely colorless llll CuNH342 CdNH342 NiNH362 CrNH35C2 C0NH35C2 Answer CdNH342 Explanation Zinc and Cadmium both have 10 electrons in the d orbital Since the dorbital is full the electrons cannot jump and produce colors The transitional metal center will have lots of ligands surrounding it Ligand an ion or molecule attached to a metal atom by coordinate bonding Cd2 will have 4dquot10 Cu 3dquot9 Ni2 3dquot8 C03 3dquot6 Colorless dquot0 and dquot10 Color dquot1dquot9 The color of a transition metal compound is determined by 0 we have to know the charge of the transitional metal complex 0 we have to know the charge of the ligands attached to the transitional metal complex The absorption spectrum of CoH2062 is given below a l l r Gamma 1 E E 11 F a l m 1 a ll Ml f V 35 15 55 ESE E39Elfl EEG wavelength rim EEIII nm EIEIEI n 4 nm 5339 m EEIII nm 430 nm 45 nm What is the color of the solution Violet 400425 nm Blue 425492 nm Green 492575 nm Yellow 575585 nm Orange 585647 nm Red 647700 nm A red B orange C yellow D green E blue F violet Answer red Explanation The green is absorbed so cannot counteract the red thus the light will be red Charge transfer also affects color dgt d transitions give color in transitional metal complexes but they are not the only reaction a transition metal complex exhibits color diffuse and re ect spectroscopy I IE39IDEHIEIZEJEJ J 532an EU m g lltrust Elf llama EEEFi lil l quot l A chemist prepared the complex ions CONH362 ZnNH362 NiNH362and collected their UV Vis spectrum Based on the results obtained from these solutions match the proper complex ion with its respective pattern Ailzrsamamice HE SQME HEE 39Ir 350 45B 553 558 SE WHVEIE EtI l f t r l 1 B 253 39 i I I I I l 253 3Cquot 3581 43E 453 5C SEE EEC BEND TD hamrhanee Wavellengm nammeters 2511 3E1 353 ml 115E SEN 55E EEIIZI wavelength nannmgtar E15111 i l l l Violet 400425 nm Blue 425492 nm Green 492575 nm Yellow 575585 nm Orange 585647 nm Red 647700 nm E D CoNH3612C znNH362 A NiNH362B l Prelecture 3541315 Chemistry of Coordination Complexes The relative energies of the dorbitals on a transition metal lead to some unique physical properties including color and magnetism The color of a complex provides insight into the structure and bonding of a materialsolution Coordination compound typically consists of a complex ion and counter ions Metallurgythe science and technology of extracting metals from their natural sources and preparing them for practical use Table 232 Properties of the Period 4 39ll39iranzaiti n Metals Group BB 413 SB 5E FE EB 13 23 Element SI Ti V Cr Mn Fe En Ni Cu in mundsmt E lecirwnEl n gmti n ErilldllSg sail53 Sliilhll g 3 Ii54l5 lial52 saw3 midi33 Mid53 MIME Ellie Firsl imiizw l inn en i gquot Ille In all 6311 4558 6 51 155 3 F1 F 755 9 E53 3 ELIE 9 DIE Metallic radius LIE4 1L4 L35 129 113 LEE 135 125 HES 137 newly gim in 45 31 quotm 22 quot39 as 351 39 Tall Ml Eltiil lg pm It GilH l 541 ll El H913quot TEE T 2114 153 494 U4 55 l lr 1 2E1 Krystal 5 tliucture hcp l l cpl b cc lam L loci hcp ft f0 11 cpl quotquot quotlhlrreivimlm1s for crystal structures nr chtp hml1gorlalclusc plucked fcc I face tentemdcuhic hot hotlymnlcmduihit arxsl Settllm Mil Mungaumisu has Illmfi CHIRPIE crystal strucl uuru Lanthanide contraction the lling of 4f orbitals through the lanthanide elements which causes a steady increase in the effective nuclear charge producing a size decrease 7 7 M j if 7 o The charge of the complex ion is determined by the charge of the counter ions 0 Ion shown above is bound directly to 6 NH3 and has an octahedral geometry 0 Everything inside the brackets will be directly bonded to the transition metal 0 Inside of brackets will act as cation 0 Outside of brackets will act as anion a I I Faramagnetic aplna ranclam apina clc align mi in a magnetic eld all l l l l l l l Ferrelma netic spina aligned parallel tn each c her It it ll l Antifermmagnetic Elana l7 Farrimagnetic unequal againa ll align in crippcaaita clircc lmczna align in cippcalita dlirecljcna but cch an cancel each lather ncit CDleEf39LElF cancel each ether 1 is El Magnetic moment a property that causes the electron to behave like a tiny magnet Diamagnetic one in which all the electrons in the solid are paired o the spinup and spindown electrons cancel one another 0 is very weakly repelled by magnets paramagnetic a substance in which the atoms or ions have one or more unpaired electrons o the electrons on one atomion do not in uence the unpaired electrons on neighboring atomsions o attracted to a magnetic eld 0 the magnetic moments tend to align parallel to one another producing a net attractive interaction with the magnet ferromagnetism when the unpaired electrons of the atomsions in a solid are in uenced by the orientations of the electrons in neighboring atomsions r permanent magnet when a ferromagnet is removed from an external magnetic eld the interactions between the electrons cause the ferromagnetic substance to maintain a magnetic moment antiferromagnetism the unpaired electrons on a given atom or ion align so that their spins are oriented in the direction opposite the spin direction on neighboring atoms 0 net magnetic moments of the spinup electrons are canceled by the spindown electrons ferrimagnetism magnetic centers have different numbers of electrons of unpaired electrons andor the number of magnetic sites aligned in one direction is larger than the number aligned in the other direction unpaired electrons align so the spins in adjacent atomsions point in opposite directions 0 the net magnetic moments of the spinup electrons are not fully canceled by the spindown electrons properties are similar to those of ferromagnetic materials Ferromagnets ferrimagnets and antiferromagnets all become paramagnetic when heated above a critical temperature 0 This occurs when the thermal energy is sufficient to overcome the forces determining the spin directions of the electrons o Called Tc Curie temperature for ferromagnets and ferrimagnets o Called TN Neel temperature for antiferromagnets Metal complexes species that are assemblies of a central transitionmetal ion bonded to a group of surrounding molecules orions Coordination compounds compounds that contain complexes Ligands the molecules or ions that bond to the metal ion in a complex 0 Functions as a lewis base and donates a pair of electrons to form the ligandmetal bond Primary valence the oxidation state of the metal Secondary valence the number of atoms bonded to the metal ion aka coordination numer Coordination sphere the central metal and the ligands bound to it Cis form the form in which the two chloride ligands occupy adjacent vertices of the octahedral arrangement Trans form the two chlorides occupy opposite vertices of the octahedral arrangement 3 J1 Two C can j Twit Cl an 0 same siclc til GPPEEJTE E l metal ion i l sides at 3x 1 5 1 metal i n E13 hammer v trams isomer counterion a noncoordinating ion that balances charge in the compound 3 Ji ll A 13 Sill if ll 3 2i ii 2 iiquot i 139 P5 quot 3 1 ll al 439 39 llCHIIIT TPI35fll32 tlximwigigt11iie 3 donor atom the ligand atom that binds to the central metal ion in a coordination complex monodentate ligands ligands having only one donor atom bidentate ligandsligands with two donor atoms polydentate ligands with three or more donor atoms chelating agents bidentate and polytenate because they appear to grasp the metal between two or more donor atoms because chele in greek means claw Lecture 3541315 Transition metals exist with a wide variety of oxidation states Indicate the coordination number and oxidation state of the metal in CoNH34C2Cl A6 2 B 7 2 C 4 2 D 6 3 E 7 3 F 4 3 AnswerD Indicate the coordination number and oxidation state of the metal in NH4CFNH32NCS4 linkage isomers Common ones are SCN and N02 0 Linkage isomers can bond with either side of the molecule because of the layout of its lone pairs When solid samples are analyzed re ectance patterns are calectecl ih stead elf absarhuahce patterns The if Visible abserptien spectrum at EtlHEDlI HI is sheath belaw I15 the ISM lE ahearaanee a1 an l l l ass as sea 55 area ass waselentn nitll lli this cemplen is crystallized sketch the re ectance pattern that waulcl be t il dn Your respnse Ln d Bi i as a harmante as M 2 5 9 59 starfl 5amp2 E5513 5 3159 wm nlgh Emnalu Prelecture 3641515 Chelate effect the trend of larger formation constants for bidentate and polydentate ligands Seguestering agents a chelating agent used to prevent one or more of the customary reaction of a metal ion without removing the ion from solution Globular protein a protein that folds into a compact roughly sphe calshape Conjugatedalternating Chelation the formation of a complex ion between a metal ion and a chelate 1 In naming eempleites that are salts the name ill the eatiem is given hefei39e name efthe I39le lllL Thus ltlelN H i ll lj 1th name the tleilth ECljlll eat ieh anti then the til i In naming eemplee inns er ti39ieleeiilesi the ligands are named hefere the metal Ligands are listed in alphabetieel melee regardlees if their eharges Prefixes that glee the mimher elligande are net eehsieleeeel part et39 the ligand name he eleteemhing alphabetical enlist Thus the Cullilllale llgm ltjlll is j ti ltitl lll lll l ll t ull lllllll Be earelill te mete heweser dial the metal is written eet in the ehehiieal tht liigulaJ The names at aniehie ligands end in the letter he hat eleeteieallle neutral li gands eetlimarily heat the meme til the meleeules i Table 235 Special Dailies are used the Ilg aqua HHj f lli t ll ll tE ul and CG earhettyll Fer example lFel l llllN39Hjlllllg jgllin is tile diammiheelaqiieelieyaheimtnlill lentil ISteel elk 1 tetra5 PEHl i here are used tn i eliieate the t l 39 ef eeeli etligahell when mere then ne is present llf the ligand E l l a Greek prefix the example E lltjllet Etll 39t letE ee pelledehtate the alternate prefixes eel hes l tetrehis al herewith head heeelris are usee ll and the ligand name is planted in herea theses Fur example the IllEEl 39tt let lCreleiilglEh is hisIEethyleiietliahiiiieleehaltIEIIIll lj39l39lill39lllldtl If the eemplee is an rantem its name ends in were The eempeuud petassiiiiii ll l tl lf ll f l l l llll the example and the lDIlL Ee ldlim te tiraelilelmeehaltate l lt ll i eeielatle member if the metal is given in parentheses in Re man numerals fellewfi g the ame f the metal Three eeah iples the applying these tales are H il NHala Ellquot leatenlginenhance DU 1 HJEE t tt t tl lt l 39t lltlltelllll l htemide eliqurElti yl llll ehisl eth elehediaiiiihelee haltl Ill i ehle title Eedium tettaehlee eeaemelyhdatelli lv l Structural isomers isomers which have different bonds 0 Linkage isomerism when a particular ligand is capable of coordinating to a metal in two ways Coordinationsphere isomers isomers that differ in which species in the complex are ligands in which are outside the coordination sphere in the solid Stereoisomersisomers which have the same bonds but different ways in which the ligands occupy the space around the metal center Geometric isomerism the arrangement of the atoms is different but the same bonds are present o Cis isomer where the same ligands are adjacent to each other 0 Trans isomer where the same ligands are opposite each other 0 Optical isomerismaka enantiomersmirror images that cannot be superimposed on each other 0 Dextrorotatory the isomer that that rotates the plane of polarization to the right Latin dexterright o Levorotatory an isomer that rotates the plane of polarization to the left Latin laevusleft Chiralmolecules or ions that are not superimposable on their mirror image 0 They have an effect on planepolarized light so are said to be optically active Solve The trans ismiier of itiuieiilg lgquot Ellid its iiiirmi iiiiiige are Mirmr ff 7 39 Ilvllkhiulu a I l i timinge Lilia Hilll 1iLLITill139iII 1 gE Elf ii isu i r is identical it the original Cuiiseqiici itly airmanam Cnier ll lg i dines not exhibit nuptial isnmcrism This iiiirmr ilimgi if the iiii isniiipr cannot bi siflperliipnscdl m1 the original Mirriir if ll n aiiiiiiiivi JEN93931 AH nggi39ahrliiilli39EJCI C flirt l id39l llll illlllfN if Cu Iquot 3 n FED if H Iir j I if If Nummmimm g ll M ll racemic mixture 0 Does not rotate polarized light because the rotatory effects of the two isomers cancel each other Polarizing U lmlaf 139d film Angle of 39 mta tier of plane all polarization Rotated polarized light l It I Upliicall active Fnl ger P la39 EEd solution a light if E m Tami 3 x a 3 Cfquot K M s a w quottil v haul 93M 3L qr 7 5 3 4 thT IN airlift 9 95quot t f quot 1 its l Elf ruff uh 39 39 Coordination Sphere the central atom and the ligands directly bound to it Coordination number the number of ligands directly connected to the transition metal center 7 1 ng if Elfm 3 Mall v39lEP a E r L Lecture 35 41515 MgS crystallizes in the NaCl structure while CdS adopts the wurtzite structure The structures and their corresponding XRay diffraction patterns are shown below E E I g l a l m E if a E E ll H 6 u 1quot i 39 4 ii E hi 3 T 7 l am 32 5 a H a E Equot E a E 53 151 Fla 4 5a 33 so Eill iIia nmea Elm rmer Several samples were synthesized and examined by powder X ray diffraction to determine if solid solutions of the type Mg1xCde formed In a paragraph tell the reader what the results of your XRD data are Identify what the composition of each sample product is The terms quotcrystalphasequot or just quotphasequot are used to talk about different crystal phases note this is not referring to state of matter solid liquid gas Was there only one crystalphase identi ed 1 Yes gt Tell the reader the identi ed crystalphase and use its chemical formula You may label this as quotphase purequot or quotsingle phasequot in referenceing it 2 No gt If you have something else present tell the reader each phase that is present in that sample Refer to the XRD gure and mention that it was noticed with additional XRD peaks present 3 Was this the expected product or something else The collected XRD patterns are shown below my 1 r WEE a agl gi I Ew 1 EL E a E 2 U 3 a1 i 3 u L of q quot Eli 51 II Mi h E 39 El iii Tn H l i m 33 l 11 3 E 39i I Fin Eti E rZi lit LEE EU iii gm 339 ign EU El nEmma EEi viimgrr Sfi V aim 5 i Eli n 53 m 2 a if 55 n II lam H a 5 Ti H in 391 E r E 1391quot 539 53 n quotj til 3 i 5 5 L at E u I it an E 3 LEE 7 Q In L 7 I3 L it E i E f E rt 3 fl 3 3 s it i 35 1 quot5 W tamermi ELF Eltienreeali An swe r 0 rst pattern is Mgo95Cdo055 Cd is doped in the M95 structure it has the rock salt pattern based on the XRD pattern it has 2 peaks in the 2040degree range 0 second pattern is Mgo80Cd02OS XRD pattern has 6 peaks 4 peaks will be from the CDs and 2peaks will be from the M95 this structure is not pure 0 third pattern is Mgo2oCdosoS XRD pattern has 5 peaks this structure is not pure fourth pattern is M8005Cdo955 XRD pattern has 4 peaks it has the structure of CdS and is pure Solid solutiona solid mixture containing a minor component uniformly distributed within the crystal lattice of the major component Tabit am Hemp1H E39ILIWn lFrutr Hgi i HE39Sgr llw m 39a I 39f EH I 3mm u rl I39Ej E Isa l i alfw l39 main t39i thaw1mg IziS I 31 Tbilag nunln39 u l39 ti til magnum Lr L5 or quot 9mth j 1quot 39 7 39 l ft 7 9939 in 33 D an m t H Dia 5nn 55n u quot H5 at 39 IJE LII39LELI uquot f39 TE 1quot f39 H fl it fl l n Finn I F Unit C 39J Vnu39u Ln u39J luau gamut Inn DI nungun L q F c 13 EH 3quot DEquot inn E ii 2 7 J 47151 4557 r m Eahrbuinurlwia hanI a E r m ESElit a eiu hiaiu ag IZ39I IEJEI aggr aftma H fara f5 h g Infhath l39Li39fEE39TJ E E39 j Elle nn39 rant c 1 PM my IIt liii2ii39 ilquot EIIE El W Hi5 Lquot39i i l l 39 33 D 2 1 5 it 9 nf lu39mmita39rWau liefE39s aptanal n fa ij r quot 12 tn tug 7 Iii u nIu llD LB 53 EELS Ellay digest cranesme L39i Sign 1 I514 abby EJEI I l ll39li39TItifIFFETlg39 EI EU 31355 i Consider the following equations NiH2062 30 6 NH3 ag F NiNH362 aci 6 H20 I l NiH2062 aq 3 en aq Nien32 aq 6 H20 I 2 where en the bidentate ligand ethylene diamine NH2CH2CH2NH2 Which of the following statements isare true A The equilibrium constant for Equation 2 K2 is larger than the equilibrium constant for Equation 1 K1 B The entropy change for Equation 1 A51 is more positive than the entropy change for Equation 2 A52 C The addition of heat to both equations favors the formation of the products You may select zero one or many answers Answer B false A true C false the higher the heat the higher the A H and the less spontaneous it will be Explanation 0 rst reaction has a AS of about 0 so the Kf will be driven by enthalpy Second reaction has a AS that will be positive The Kf will be driven by enthalpy and entropy Prelecture 3641715 Spectrochemical series ligands are arranged in the order of their abilities to increase splitting energy as in the following list increasin it at ICIT lFT HEB MHZ en NGETrEN lmnded ECH spinpairing energy the difference between the energy needed to pair an electron in an occupied orbital and the energy required to place that electron in an empty orbital This is caused by the fact that electrostatic repulsion between two electrons that share an orbitaland have opposite spins is greater than that of two electrons that are in different orbitals with parallel spins Hidhspin complex a complex in which the electrons are arranged so that they remain unpaired as much as possible Lowspin complex a complex in which the electrons are arranged so that they remain paired as much as possible which still following Hund s rule 0 all tetrahedral complexes are high spin The crystal eld splitting energy is never large enough to overcome the spinpairing energies squareplanar complexes are nearly always low spin with the 8 d electrons spinpaired to form a diamagnetic complex leaving the dXZy2 orbital empty Crystal eld splitting energyA the energy difference between the sets of d orbitals Strong eld ligand a ligand with a large A Weak eld ligand a ligand with a small A 35 it hplanck s constant 663E34J s C speed of light 300E8 ms E splitting energyA f a I ll T Tryatal aa 31 15511 3 quot a a p ll il ting quot a a a 39 l E d I I aquot 1339 y II II Metal inn in an If 1 itquot E tiil lELlTLill cumpla is f II in Elfluff E u39 J P Ir F f 39 a f if f39ia letal inn in a i 39f aq nare planar crumplea Free metal iian Free metal inn 13quot I a l E a pr 4 A Natal inn in a tetrahedral unnple Free 111atal inn Lecture 36 41715 A molecule forms to stabilize the energy of the atoms involved putting electrons in low energy orbitals The greater the orbital overlap the stronger the bond the lower the energy The greater the orbital overlap the stronger the antibond the higher the energy If the overlap is the same the energy will be the same Degenerate orbitalsorbitals of the same energy dictated by the geometry large A for octahedral complexes with CO or CN ligands or transition metals in the 4d or 5d area 2 axis of a tetrahedron cuts straight through the molecular bonds so the four ligands will not interact in the same way A t 49 A 0 Magnetic measurements were made on the following ve complexes which ones were found to be paramagnetic A MnBr4 B PdCI42 C VNH365 D FeCO62 E COCN63 AnswerA On square planar complexes most are d8 compounds 0 with these compounds there are zero unpaired electrons they are diamagnetic with strong eld ligands the octahedrons are diamagnetic with weak eld ligands the octahedrons are paramagnetic Lecture 42015 MnenzBr2 can exist as a cis and a trans isomer Which of the cis and trans isomers can have optical isomers You are not sitting close enough to other students for us to automatically nd a group for you Please nd a student sitting nearby and discuss your response with them A cis only B trans only C both cis and trans D neither cis nor trans E anen2Br2l does not exhibit cistransisomerism Answer A Explanation Trans isomer does not have an optical isomer Cis cannot be superimposed on its mirror image Thus it is an optical isomer 0 Optical activity is when the mirror image of a complex cannot be superimposed on its mirror image What type of isomers exist for Coen2C2 A linkage B coordinationsphere C geometric D optical You may select zero one or many answers Answer C D Explanation In a linkage isomer one of the ligands has to bond differentlyex SCN orNCSL In a coordination sphere isomer the salt center ion quotswitches partnersquot in the coordination compound one of the ions bonded switches and becomes the counterion and the counterion becomes the ion in the coordination compound Geometric cistrans Optical nonsuperimposable mirror images The correct arrangement of the following complex ions in terms of increasing crystal eld splitting energy A isCr6339 CrF439 Cren33 Wen33 CrI63 CrF4 Cren33 Wen33 lt CrI63 CrF4 Cren33 Wen33 lt CrI63 CrF4 Cren33 Wen33 lt CrI63 CrF4 Cren33 Wen33 Prerecitation 4212015 Ftroolem 2352 As shown in Figure 1 the lot d tiransition of TiIng Ull l lll produces an absorption maximum at a wavelength of about Elli Jim I Figure l Iof1 Blue greenqeilow absorbed a mixture of violet and red light trend to the eye soluljon E39F F EE E PUFF39lE Absentbones Part A What is the magnitude of y for TIiliH2 3i in halfmo ll 239 Millimot My Answers Lip Correct l Part B How would the magnitude of in change it the Hg C ligands in Tifing Ull l ln were replaced with NH ligands a The magnitude of in would increase The magnitude of u would decrease 39lhe magnitude ol it would not change My Answers Give LI o Eorrect Provide Feedback Continue lncneast1fig t quot 21 s F s H s hing s an s MATH localised s IN 400 450 500 550 600 650 700 Wavelength nm Eshv mili lsquot l E illlll lll l llpliol m l 23 ll 5 Prelecture 3842115 Nuclear reactions transformations of atomic nuclei Nucleons subatomic particles that reside in the nucleus protons and neutrons Atomic number number of protons Mass number total number of nucleons in the nucleus lsotopes atoms with the same atomic number but different mass numbers heirs These isethpes ere else written QED and 2331 where the superserlpt is the mess number and the subseript is the etet39nie number Nuclide a nucleus containing a specifies number of protons and neutrons Radionuclides a nuclide that is radioactive Radioisotopes atoms containing radionuclides Alpha particles helium4 particles Alpha radiation a stream of alpha particles Example of nuclear equation lager v H 33th isle Alpha decay when a radioactive element loses an alpha particle like in the equation above Beta particles electrons emitted by unstable nucleus 39E o The superscript 0 represents that the mass of the beta particle is negligible compared to the nucleons Beta emission iiiI iii s ie 0 It39s equivalent to the conversion of a neutron to a proton Gamma radiation high energy photons doesn39t change atomic number or n a 7 V mass number represented by Uquot m 5511111311 ii Positron emission 1amp1 is fe Has the effect of converting a proton to a neutron Electron capture the capture by the nucleus of an electron from the electron cloud surrounding the nucleus le fe orbital electron gfi r Has the effect of converting a proton to a neutron Table 21 Particles Found in Nuclear Reactions Particle Symbol Neutron n or n Proton jII or p Electron Jiie iipiha particie 3171c or or Beta particle e or Positron g m7 Table E13 Types of Radioactive Eecay Mpha dec a Beta emission Positron emission Eilectron capturet Nuclear Equation ange in atomic Number a so 2 Dquot ig clji39Y 39iquot if 1 tit r f it i H gitquot 1 Strange in Mass Number ii Unchanged Unchanged Unchanged quotJ The electron captured comes from the electron cloud surrounding the nucleus tThe positron has a 39Irkerjgrr short life because it is annihilated 1when it collides with an electrom producing I n gamma rays le l e 3 Emir Radioactivity is tihe spontaneous emission of radiation from an unstable nucleus There are ve major types of radioactive decay i Aipiha or radiation consists o39l helium nuclei The helium nucleus is a small particle containing two protons and two neutrons EHo 2 Beta i3 radiation con sists of electrons 31 e 3 Positron emission results from the conversion of a proton in the nucleus to a neutron The ejected positron iiio is a particle tihat has the same mass as an electron but an opposite charg e 21 Eiectron capture is tihe capture of an inner sheii electron by a proton in the nucleus The process emits gamma or radiation and results in a proton converting to a neutron Gamma radiation consists of high energy electromagnetic radiation 5 Gamma Pg radiation consists of high energyr radiation and contain no particies and tlhus theyr have no mass 1hr Titere are two main rules to remember when baiancing nuclear equations i The totai of the superscripts mass numbers in the reactants and products must be the same 2 The totai of the subscripts atornic numbers in tihe reactants and products must be the same Mass number a I it 1 Elicnticai Syruboi Atomic number lad439 strono nuclear force a strong force of attraction between nucleons all nuclei with 84 or more protons are radioactive ll lJ jl Nuclei with E 2 e4 domii iam idem made 3 alpha emieeien 15H Mi 233 inf 4 152 llS Nuclei abeve belt at Stability clemihaht decay mecle E i beta emissieh Jill llilll 39 deminaet clecar mede E peeitren f1 emissieri er electmri capture J Nluclei belew belt ei stability Ill 39E ll Ell twp 1411 Ell 39 Number elf n eutrene quotll 1 neutrelm te preten ratiiei v ggFe ham In llH quotTurnip THU l lit 2t 3t 4U St til 27 8t Qt 39llllll Number ell preterm p it Figure 21 Stable and reclinerstitre iisetepee as a fune tien ef numbEre of neutrene and pretene in a nucleue The stable nuelei dark blue clots define a regien Hnewn as the belt at stability general scenarios of radioactive decay 1 Nuclei above the belt of stability high neutrontoproton ratios a These nuclei can lower their ratio an move toward the belt of stability by emitting a beta particle 2 Nuclei below the belt of stabilityow neutrontoproton ratios a These nuclei can increase their ratio and move closer to the belt of stability by either positron emission or electron capture positron emission is more common among lighter nuclei electron capture becomes increasingly common as the nuclear charge increases 3 Nuclei with atomic numbers greater than or equal to 84 a These nuclei tend to undergo alpha emission which decreases both the number of neutrons and number of protons by two moving the nucleus toward the belt of stability Radioactive decay chain nuclear disintegration series a series of nuclear reactions that begins with an unstable nucleus and terminates with a stable one o Nuclei with even numbers of protons neutrons or both are more likely to be stable than those with odd numbers of protonsneutrons Nuclei with 2 8 20 28 50 or 82 protons are generally more stable Nuclei with 2 8 20 28 50 82 or 126 neutrons are generally more stable 1 Number 431 l H if Mable i3tquotilupe3 HE 2 ElemerltH 3339i 3911 twn 1 err fenvet Htalnle lentnpee 5 g Li Be 3 c N In F We 2139 ll ELEE HE HIH with three 2 1 i2 3 if 3 11 39 nr r nnre stable ieullupeea 13 39 M 15 7 71E 1 N21 Al P 39539 C1 A ill 3 iii l3 ill 4i l2 3 w 33 an 33 33 33 33 33 33 33 33 33 Ca Ti 39339 E Ee Ni Ge A3 Se Er 1K3 lEi 5 ll 5 El iii ill 4 7 i5 ill i5 El 4 ill ii iii iiil 3339 33 33 33 33 33 33 33 33 33 33 33 33 31 33 33 33 Kb 51quot if Elmquot Nib MD Tc Ru Pd g Ed 111 Sin 5b TE I Pie 1 ll 3 ill 4 ill 6 iii Fl ill El i2 6 1 ill i2 6 ill 3 Lecture42115 Atoms with an atomic number of less than 70 is stable Atoms with an atomic number greater than 84 is radioactive One of the nuclides in the following pairs is radioactive Select the stable isotope from each pair Note You will be selecting 3 answers in this question one for each pair A Pair 1 209mm B Pair 1 2080833i C Pair 2 3919 D Pair 2 4019 E Pair 3 nickel58 F Pair 3 nickel65 You may select zero one or many answers Answer ACE Explanation Bi209 is stable because the number of neutrons is even K 39 is stable because the number of neutrons is even Ni58 is stable because both the number of neutrons and protons are even ill A FELLIET F FJLU TMUMEEE MIME Wi ii i i geriarats hat ari ight from its raioa tiw may Faiiata SiLiEii as the ar Liad iri raioiaotop tiirrriolaratrii amaratars HTGSiI to riarat El tri ity iri aiaz wahi iaa The plutonium238 that is shown in the chapteropening photograph undergoes alpha decay What product forms when this radionuclide decays A Plutonium234 B Uranium234 C Uranium238 D Thorium236 E Neptunium237 Answer B Explanation helium4 is emitted subtract 4 from 298 and 2 from 94 The radioactive decay of thorium232 occurs in multiple steps called a radioactive decay chain The second product produced in this chain is actinium228 Which of the following processes could lead to this product starting with thorium232 A Beta emission followed by electron capture B Alpha decay followed by beta emission C Positron emission followed by alpha decay D Electron capture followed by positron emission E More than one of the above is consistent with the observed transformation Answer B ExplanationWWff if h Mi n 39 ii i e l l l Prelecture 34 4232015 Battery a portable selfcontained electrochemical power source that consists of one or more voltaic ces Primary ces cannot be recharged and must be discarded or recycled after the voltage drops to zero Secondary cell can be recharged from an external power source after its voltage has dropped Fuel ces similar to batteries but the fuel must be continuously supplied to generate electricity Corrosion a spontaneous redox reaction in which a metal is attacked by a substance in its environment and converted to an unwanted compound Cathodic protection protecting a metal from corrosion by making it the cathode in an electrochemical cell Sacri cial anode the metal that is oxidized while protecting the cathode Electrolysis reaction a reaction that requires electricity to occur Electrolytic ce two electrodes immersed in solution and a battery nonspontaneous Inert electrodes do not react Active electrodes participate in the electrolytic process 1 mol eectrons96485 C coulombs coulomb the quantity of charge passing a point in the circuit in 1 second when the current is 1 ampere A cgulombs amperesHX seconds cm 4 4 d 9 care firmly E ifquot W a F Mills 9 Eire 0 Ergm 111 995 all 23m I rimfrag quotr and 39 W qggrag if I M5 air2339ny faint 5H1 Eh quot39 malt 5395quotquot was Eel3 Prelecture 3942315 Halflife the time required for half of a given quantity of a substance to react Radioactive decay Rate kN kdecay constant N nuclei Activity the rate at which a sample decays Becquerel Bql SI unit for activity one nuclear disintegration per second Curie Cil 37E10 disintegrations per second Nu mass defect the mass difference between a nucleus and its constituent nucleons E r32 m c3E8 nuclear bindino enerov the energy required to separate a nucleus into its individual nucleons fission where nuclei gain stability and give off energy if they are fragmented into two midsized nuclei the original nuclei are usually large fusion where very light nuclei are combined to release energy Eh Ii 2 a a e filling 2 7 L r T 3917 I 7 V r 7 r i it 39 9 v d Spitting nuclei T i ssion 3 u r 1 n m r n 7 7 rete ea vaneerg gal E v Combining uncle1quot E E lif i El releae er energy H E l ll 5i Will ill Elli E 25 E rmnfilljer ail Figure 112 lHIJElIlEEr binding energis The varage inlilin enemy pi FillJE IE l39l EWEFEEEES initially s the mass numlzar in esea and than decrease5 slowly Hemline f these trends iusin of light nuclei and fission of new nuclei ar exothermic iplrocesses Lecture 3942415 How many on and 5 emissions are involved in 238U decay 238U 206Pb Enter you answer in the space below with the x emissions first separated by a comma then the 5 emissions for example if there are 3 x emissions and 2 5 emissions you would enter 32 Cesium137 which has a halflife of 302 years is a component of the radioactive waste from nuclear power plants If the activity due to cesium137 in a sample of radioactive waste has decreased to 352 of its initial value how old is the sample A 104 yr B 154 yr C 315 yr D 455 yr E 156 yr A wooden artifact from a Chinese temple has a 14C activity of 380 counts per minute as compared with an activity of 582 counts per minute for a standard zero age From the half life of 14C decay 5715 yr determine how many years old the artifact is the larger the transition metal center the further out the delectrons go into space
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