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by: Brady Spinka


Marketplace > University of Texas at Austin > Chemistry > CH 301 > PRINCIPLES OF CHEMISTRY I
Brady Spinka
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This 29 page Class Notes was uploaded by Brady Spinka on Monday September 7, 2015. The Class Notes belongs to CH 301 at University of Texas at Austin taught by Staff in Fall. Since its upload, it has received 19 views. For similar materials see /class/181885/ch-301-university-of-texas-at-austin in Chemistry at University of Texas at Austin.

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Date Created: 09/07/15
Lecture 1 Physical Equilibria The Temperature Dependence of Vapor Pressure Our first foray into equilibria is to examine phenomena associated with two phases of matter achieving equilibrium in which the free energy in each phase is the same and there is no change in the overall values of system state functions The areas of physical equilibria we will investigate are 0 The interface between phases gas 6 9 liquids 6 9 solid Solution properties like freezing and boiling will be given a thorough thermodynamic treatment Mixtures formed when phases are soluble or miscible in one another Dissolving solids or gases in liquids or mixing two liquids Colligative properties that derive from mixing of two phases These are essential properties of biological and chemical systems We will draw on the following theoretical treatments from CH 301 o intermolecular force theory 0 thermodynamics So review these concepts Specifically review the intermolecular force and thermodynamic lecture notes and worksheets Pay special attention to the introductory treatment of phases and phase transitions that are presented at various times during these lectures Vapor Pressure Vapor pressure was introduced in the fall semester as a solution property and we learned that vapor pressure rankings could be qualitatively determined from a ranking of intermolecular forces De nition of vapor pressure the pressure exerted by the vapor of a substance that exists in a condensed liquid or solid phase Just looking at this picture suggests vapor pressure is a surface phenomenon The more surface the more vapor The depth of the condensed phase doesn t matter The reason is that the particles in the condensed phase With the greatest ability to enter the vapor phase are on the surface Waiev Vacuum up Vapor pressure IMF Theory and vapor phase ranking It is intuitive that the stronger the intermolecular force holding a substance together the less likely it is to enter the vapor phase Similarly the smaller the IMF force the greater the vapor pressure Here is the ranking and trend contrasting IMF and assorted compounds As the lMF increases Instantaneous dipoles lt1 kJmole lt dipole dipole 5 kJmole lt H bonding 20 kJmole lt ionic 200 kJmole The vapor pressure decreases He gt CH4 gt C3Hx gt CCl4 gt CHCl3 CH3OH gt H20 gt NaCl gt CaO It is pretty easy to generate these qualitative rankings of vapor pressure by first ranking IMF Look at the table of values for different compounds at 25 C to see the correlation TABLE 82 Vapor Pressures at 25 C Substance Vapor pressure Torr benzene 946 ethanol 5 8 9 mercury 00017 methanol 1227 waterquot 23 8 For values at other temperatures see Table 83 Now can we create a quantitative measure of vapor pressure First recognize that vapor pressure is an equilibrium phenomenon that is temperature dependent 6 note that at equilibrium as vapor encounters the surface it condenses and the energy released promotes a different T particle into the vapor 6 also note that at the higher T more particles achieve separation energy to enter vapor mu 83 Vapor Pressure of Water Tclnpuralure upnr pmsum Cl l I In the table of water vapor pressures note that as the temperature increases the vapor 10 pressure increases until the boiling point is reachediIOO degrees at l atmosphere Also note What appears to be an exponentiallike 3 increase in P as T increases vli 11944 525 Inn mull 39I39rmli mnpumum Can we derive an expression that predicts this relationship First note that at equilibrium A Gvap Ggas Gliquid 0 Guquid G0 liquid because it is pressure independent Ggas G gas RTlnP Where RTlnP correction for change in pressure so A Gvap G gas RTlnP G uquid G gas G uquid RTlnP A G vap RTlnP the standard free energy of vaporization But A G vap AH vap TAS vap So lnP AG m AH VE AS Vap A constant assuming ideal gas behavior RT RT R because a mole of any kind ofgas has about the same entropy Or multiplying through by eX AH vapRT P Ke This is a very famous mathematical relationship that shows up all the time in nature What does the above equation suggest for a function An exponential increase in P with T Note this functional relationship not just for water but for other compounds 100 i 150 Note that as delta H vap so D39e wlefh quot decreases With A Ethanol i 3 a increasmgIMF the E g 110 bOiling pomt increases 3 so i 7 7 a and that this correlates g g W A With IMF 5 40 g g gt 7 T 39 39 g I l I 120 90 100 1 0 Temperamre T 0 4 407100 750 a so 100 TemperatureFC And now for something really useful that scientists do Combining equations to eliminate a variable or how to create the famous Clausius Clapeyron equation Given that As vap is assumed to be constant in the expression In P AH m AS m R R 0 We can rearrange the equation to put everything in terms of Asvap identify two states of the system P1 T1 and P2 V2 set them equal and do mad algebra When it is all over In a m Li Pl R T1 T2 This is the famous ClausiusClapeyron equation that allows us to use one value of P1 and T1 to predict the vapor pressure anywhere else on the graph So let s test this by using the equation to find the boiling point of water hey at least we ll know if our answer is right since at 100 OC degrees the vapor pressure should be 760 torr Selecting a P and T from our table of H20 vapor pressures Like 20 OC and 1754 torr and given AHO vap for H20 407 kJmole We need to work in the correct units so convert 1754 torr into Pa ltorr 1333 Pa so 1754 torr 1333Paltorr 23381 Pa rearranging 0 lug AHvap ii 9 ii R 01mg P1 R T1 T2 T1 T2 AHVaP P1 8A 1 1 Perlodlc Table of the Elements 2 18 H 2A 3A 4A 5A 6A 7A He 10079 2 13 14 15 16 17 40026 3 4 5 5 7 a 9 10 Li Be B C N O F Ne 6 941 9 0122 10 811 12 011 14 0067 15 9994 18 9984 201797 11 12 13 14 15 15 17 15 Na Mg 3B 48 SB 53 7B 83 1B 2B AI SI P 8 CI Ar 22 9898 24 3050 3 4 5 e 7 B 9 1o 11 12 26 9815 28 0855 30 9738 32 066 35 4527 39 948 19 2o 21 22 23 24 25 25 27 2a 29 3o 31 32 33 34 35 35 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 39 0983 40 078 44 9559 47 88 50 9415 51 9961 54 9380 55 847 58 9332 58 69 63 546 65 39 69 723 72 61 74 9216 78 96 79 904 83 80 37 3a 39 4o 41 42 43 44 45 45 47 4a 49 5o 51 52 53 54 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te Xe 85 4678 87 62 88 9059 91224 92 9064 95 94 98 10107 102 9055 106 42 107 8682 112 411 114 82 118 710 12175 127 60 126 9045 13139 55 55 57 72 73 74 75 75 77 7a 79 50 a1 a2 a3 a4 a5 55 CS Ba La Hf Ta W Re Os Ir Pt Au Hg TI Pb BI Po At Rn 132 9054 137 327 138 9055 178 49 180 9479 183 85 186 207 190 2 192 22 195 08 196 9665 200 59 204 3833 207 2 208 9804 209 210 222 57 aa 59 104 105 105 107 105 109 Fr Ra Ac Rf Db Sg Bh Hs Mt 223 225 227 251 252 253 252 255 255 5a 59 5o 51 52 53 54 55 55 57 5a 59 7o 71 Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu 140 115 140 9076 144 24 145 150 36 151 965 157 25 158 9253 162 50 164 9303 167 26 168 9342 173 04 174 967 90 91 92 93 94 95 95 97 9a 99 100 101 102 103 Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr 232 0381 231 0359 238 0289 237 244 243 247 247 251 252 257 258 259 250 Version 001 7 Exam 1 7 David Laude 54705 2 This print out should have 30 questions Multiple choice questions may continue on the next column or page 7 nd all choices before answering V11 V21 V31 V41 V52 ACAMP 02 0001 0802 general multiple choice gt 1 min xed 001 part 1 of 1 6 points A 200 nm photon has how many times the energy of a 700 nm photon 1 35 correct 2 42 3 029 4 024 5 993 X 10 19 6 284 X 1019 Explanation Energy of Light E E For the 200 nm photon he E 7 7 6626 X 1034 J s3 X 108 m s71 200 X 109 m 994 X 1019 J For the 700 nm photon he E A 7 6626 X 1034 J s3 X 108 m s71 700 X 109 m 284 X 10 19 J Thus 994 X 1019 J 35 284 X 10719 J ChemPrin3e T01 26 0805 general multiple choice lt 1 min xed 002 part 1 of 1 6 points Which of the following emission lines corre sponds to part of the Balmer series of lines in the spectrum of a hydrogen atom A n 2 7 n 1 B C 7174771 D n737n2 E 7174771 1 B and D only correct 2 A D and E only A A and C only P E only 9quot B and C only 9 D and E only 7 B C and E only Explanation The Balmer series is produced by elec tronic transitions which either begin absorp tion spectra or end emission spectra at the energy level n 2 These correspond mostly to the visible region ChemPrin3e 01 30 0805 general numeric gt 1 min normal 003 part 1 of 1 6 points In the spectrum of atomic hydrogen a vio let line is observed at 434 nm What are the beginning and ending energy levels of the elec tron during the emission of energy that leads to this spectral line 1 n 5 n 2 correct 2 n6n2 3 n6n3 4 n5n3 5 n4n2 Version 001 7 Exam 1 7 David Laude 54705 3 6 n 4 n 3 Explanation 434 nm 434 gtlt10 7 m Because the line is in the visible part of the spectrum it belongs to the Balmer series for which the ending 71 is 2 For the starting value of n c 3 X 108 ms V7X7434gtlt10 7m 6909 X 1014 s71 Using the Ryberg formula 1 1 1329gtlt1015S 1ltF7 2gt 2 1 7 1 1 329 X 1015 s 1 7 n7 7 Kg 1 7 1 1 Kg 7 n7 7 329 gtlt1015 s 1 1 6909 X 1014 s71 4 7 329 X 1015 s 1 0 OneD Ground State 0806 general multiple choice lt 1 min xed 004 part 1 of 1 6 points lf a particle is in a one dimensional box and is in its ground state where would you MOST probably nd the particle 1 in the center of the box correct 2 at the two ends of the box 3 either side of the center of the box 4 anywhere in the box Explanation ChemPrin3e T01 78 0803 general multiple choice lt 1 min xed 005 part 1 of 1 6 points Estimate the minimum uncertainty in the po sition of an electron of mass 9109 X 10731 kg if the error in its speed is 300000 ms 1 386 pm 2 386 X 10 12 m 3 193 pm correct 4 193 X 10 12 m Explanation m 9109 X 10 31 kg Av 1300000 ms 1055 X 1034 J s 2 9109 X 1031 kg 300000 ms 1012 193033 X 1010 m 1m 193033 pm DeBroglie Wavelength 03 0803 general multiple choice lt 1 min xed 006 part 1 of 1 6 points What is the de Broglie wavelength of Schrodinger7s cat Albert running to his food bowl Albert has a mass of 5200 g and is running at 16 ms 1 7964 X 10735 m correct 2 7964 X 10 38 m 3 4978 X 1035 m 4 5513 X 10 33 m Explanation A E h p m 1 i 6626 X 1034 52 kg 16 ms 796394 X 10 35 m Version 001 7 Exam 1 7 David Laude 54705 4 Schrodinger Eq 01 0806 general multiple choice lt 1 min xed 007 part 1 of 1 6 points Which of the following applications of the Schrodinger equation includes a potential en ergy term With both attractive and repulsive terms 1 Vr for electrons in the helium atom correct 2 V1 for a particle in a box 3 Vr for the electron in the hydrogen atom 4 V1 for the standing wave of a plucked guitar string 5 None of these Explanation ChemPrin3e T01 38 0807 general multiple choice lt 1 min xed 008 part 1 of 1 6 points The three quantum numbers for an electron in a hydrogen atom in a certain state are n 4 6 1 mg 1 The electron is located in What type of orbital 1 4s 2 3p 3 3d 4 4d 5 4 correct Explanation The notation is m Where n 1gt2gt3gt4gt5gtmgt 0gt1gt2gtmgtltn 71 represented as a letter 6 0 7 s 6 1 7 p 627d637gtf647gtg6 5 7 h etc and my 76 76 71 76 7 2gt gt 0gt gt T 2gt 7 1gt The value of mg is not needed to determine the orbital type as long as it is valid Quantum Number 01 0808 general multiple choice lt 1 min xed 009 part 10f 1 6 points What is the total number of orbitals found in the n 1 through 71 4 shells 1 30 correct 2 16 3 60 4 10 5 None of these Explanation Degenerate Energy Levels 08 12 general multiple choice lt 1 min xed 010 part 10f 1 6 points Which of the subshells possess degenerate en ergy levels 1 All except for s subshells correct 2 Only 3 subshells 3 Only subshells found in the periodic ta ble 4 Only the 2p subshells 5 All of them Explanation Electron Con g 01 08 13 general multiple choice lt 1 min xed 011 part 10f 1 6 points Which of the following ions or neutral atoms does NOT possess the electronic con guration Ar 4323032 1 V7 correct 2 Fe4 Version 001 7 Exam 1 7 David Laude 54705 5 3 Ti 4 H21 5 Ca27 Explanation ChemPrin3e T01 41 0812 general multiple choice lt 1 min xed 012 part 1 of 1 6 points Write the ground state electron con guration of a chromium atom 1 Ar 4523014 2 Ar 4913015 3 Ar 3035 431 correct 4 Arl3d6 5 Arl3d4432 Explanation The Aufbau order of electron lling is 13 23 2 33 3 43 3d 4 53 4d 5 63 4f 5d 6 etc 3 orbitals can hold 2 electrons p orbitals 6 electrons and d orbitals 10 electrons Note some exceptions do occur in the electron con guration of atoms because of the stability of either a full or half full outermost d orbital so you may need to account for this by shu ling7 an electron from the n 7 1 3 orbital Finally use noble gas shorthand to get the answer Ar 3035 431 Valence Con g 01 0813 general multiple choice lt 1 min xed 013 part 1 of 1 6 points Which of the following valence con gurations is INCORRECT 1 Tl2 32d10 correct 2 Cu 31d10 3 Bi2 32dlop1 4 3105 d10 5 Pt 110 Explanation Periodic Table 01 0811 general multiple choice lt 1 min xed 014 part 1 of 1 6 points Which of the following statements that de scribe the periodic table is true 1 The rst family of elements on the peri odic table is the alkali metals correct 2 Rows of the periodic table are referred to as families 3 An element with an 32135 valence shell is a noble gas 4 There are three blocks represented on the periodic table of the elements 5 The main group elements are found in the 3 p and d blocks Explanation ACAMP304 E2 01 0903 general multiple choice lt 1 min xed 015 part 1 of 1 6 points Why is it harder to remove an electron from uorine than from carbon or to put it an other way why are the valence electrons of uorine more strongly bound than those of carbon 1 Fluorine has more valence elctrons than does carbon 2 Carbon has a lower atomic mass than does uorine 3 The valence electrons of both uorine and carbon are found at about the same distance from their respective nuclei but the greater positive charge of the uorine nucleus attracts its valence electrons more strongly correct Version 001 7 Exam 1 7 David Laude 54705 6 4 The statement is false it takes very nearly the same energy to remove an electron from ionize both elements 5 Fluorine has a nearly lled octet which is always more stable than a partially lled octet Explanation For each element the valence electrons are in the 2s and the 2p orbitals C 1522522p2 6 protons in the nucleus 6 electrons total 4 valence electrons F 1522522p5 9 protons in the nucleus 9 electrons total 7 valence electrons The effective nuclear charge felt by the valence electrons of uorine is greater than the effective nuclear charge felt by the va lence electrons of carbon Therefore the va lence electrons of uorine experience a greater Coulombic attractive force and are harder to remove Msci 06 0201 0901 general multiple choice gt 1 min xed 016 part 1 of 1 6 points For which of the following elements would the size of the neutral atom atomic radius be largest 1 Rb correct 2 K 3 Na Explanation Atomic radii become larger as you move from right to left across a row and also larger as you move down a column Diagonal rela tionships can be tricky especially when you have to decide which of the two relationships will be the most important Here luckily the comparion works well With each successive member of a column you are introducing a new energy level farther and farther from the nucleus The largest radius here would then belong to the element which sits closest to the bottom left corner of the periodic table which is Rb in this example Ionization Energy 0903 general multiple choice lt 1 min xed 017 part 1 of 1 6 points Rank Na Mg Al and Si in terms of increasing ionization energy 1 Na lt Al lt Mg lt Si correct 2NaltMgltAlltSi 3SiltMgltAlltNa 4SiltAlltMgltNa 5NaltSiltMgltAl Explanation Mlib 07 0225 1102 general multiple choice gt 1 min xed 018 part 1 of 1 6 points Which of the following is the best represen tation of the compound potassium sul de 1 2K S correct 2K22S 3 K S I 27 4 K2 S I 37 53K2 2 z s z 6 None is appropriate because potassium sul de is a covalent compound Explanation The best drawing will show the valence elec Version 001 7 Exam 1 7 David Laude 54705 7 trons the charges and the appropriate ratio of ions for K28 ChemPrin3e T02 07 1005 general multiple choice lt 1 min xed 019 part 1 of 1 6 points Which of the following has the highest lattice energy 1 NaCl 2 K1 3 MgO correct 4 BaO 5 CaO Explanation Mg2 and 02 have the highest charge den sities DAL 07 013 1107 general multiple choice gt 1 min xed 020 part 1 of 1 6 points In the following group of compounds ogsogsogig0 identify those that exhibit resonance 1 All exhibit resonance 2 03 802 803 C052 only correct 3soamp1gco onw 4 C052 only 5 802 803 only Explanation All except 1 exhibit resonance Resonance occurs when more than one structure can be drawn for a compound involving changing the position of double bonds 03 has 2 resonance structures 802 has 2 resonance structures 80 has 3 resonance structures and C052 has 3 resonance structures 1 has 2 single bonds and no double bonds there is no resonance for this structure ChemPrin3e T02 22 1104 basic multiple choice lt 1 min xed 021 part 1 of 1 6 points How many lone pairs of electrons are found in the Lewis structure of hydrazine HgNNHg 1 8 2 4 3 1 4 0 5 2 correct Explanation H N N H The Lewis structure is H H ChemPrin3e 02 52 1109 general multiple choice lt 1 min xed 022 part 1 of 1 6 points Which of the species NO BrO are radicals ong BF 1 NO and BrO only correct 2NOamiCH omy 3 BrO and CH only 4 BrO and B134 only 5 NO CH and BF only 6BrOCH andBFonw 7 NO and BF only Explanation The Lewis structures are N 6 13r 6 Version 001 7 Exam 1 7 David Laude 54705 8 15 7 39 39 H J H 8 fl 3F B F3 o I Cl B 5 Cl Radicals are species with an unpaired elec tron so only NO and BrO are radicals 9 CICIEBCI Lewis BC13 dash 1105 general multiple choice lt 1 min xed ClB 023 part 1 of 1 6 points 10 Which of the following is the correct Lewis Cl 39 C formula for boron trichloride BClg Explanation Cl B Cl B contributes 3 valence 6 and each Cl 1 contributes 7 valence 6 for a total of 24 6 i Cl i B is a known exception to the octet rule and quot can form stable with 6 valence 6 Cl B Cl quot quot 2 l correct i Cl B Cl 31 Cl Cl JJL 7040110 quot 1104 general multiple choice gt 1 min 3 Cl B 39 H wording variable 39 Clquot 024 part 10f 1 6 points H H Which of these substances has a Lewis for Cl B Cl mula incorporating a double bond 4 l Cl 1 C2H4 correct 2 HBr 5 Cl B Cl 3 cs C1 quot 4 HgTe 6 ClJI3 Cl 5513143 31 6 HCl 7 PHg 7 Cl B Cl I 8 HF Cl Version 001 7 Exam 1 7 David Laude 54705 9 10 Ang Explanation The structure for C2H4 is H H C C H H Each C has 4 valence 6 and each H has 1 for a total of 12 6 Mlib 76 1067 1106 general multiple choice gt 1 min xed 025 part 1 of 1 6 points Which one of the following compounds does NOT obey the octet rule 1 SF6 correct 2 H20 3 CH4 4 NH3 Explanation S has 6 valence 6 F has 6 X 7 valence 6 Total 48 valence 6 h s r39 SF6 has 48 6 but S has an expanded valence shell containing 12 6 thus violating the octet rule Lone Pairs 02 1106 general multiple choice lt 1 min xed 026 part 1 of 1 6 points How many lone electron pairs are found on the central atom of ng 1 3 correct 21 32 44 55 Explanation Msci 15 0006 1108 general multiple choice gt 1 min wording variable 027 part 1 of 1 6 points For the electron dot representation d 1 Q N 210 03 3 of N205 What are the formal charges on each atom going from 1 to 4 in order joquot 4 N 1 1 0 71 1 correct 2 0 0 1 71 3 0 0 71 0 4 1 0 0 71 5 0 0 0 0 6 71 2 1 0 Explanation The formal charge is calculated by PC group 7 bonds unshared 6 Thus F0157401 FCQ67240 FCg6i1671 F0457401 ChemPrin3e T02 30 1108 general multiple choice lt 1 min xed 028 part 1 of 1 6 points Write three Lewis structures for the cyanate ion NCO Where the arrangement of atoms Version 001 7 Exam 1 7 David Laude 54705 10 is N C and 0 respectively In the most 4 H Si plausible structure using formal charges 5 H Cl 1 there is a triple bond between N and C E xplanatlon correct 2 there are two double bonds 3 there is a triple bond between C and O 4 the formal charge on O is 1 5 the formal charge on N is 71 Explanation 0 0 1 2 N E C O 2 1 0 9 g 0 1 NCO 2N CEO Bond Lengths 1009 general multiple choice lt 1 min xed 029 part 1 of 1 6 points Using polarizabilty arguments rank the bond lengths from longest to shortest for the H atom attached to a halogen atom 1 Hi gt HBr gt HCl gt HF correct 2 H1 gt HCl gt HF gt HBr 3 HF gt HCl gt HBr gt H1 4 H01 gt HF gt H1 gt HBr Explanation Electronegativity Diff 01 0905 general multiple choice lt 1 min xed 030 part 1 of 1 6 points For which of the following bonds is the differ ence in electronegatiVity the largest 1 H F correct 2 H H 3 H C gt050 Periodic Table of the Elements 58 59 60 62 64 65 66 67 71 Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu 1401 1409 1442 145 1504 1520 1573 1509 1625 1649 1673 1609 1730 1750 90 91 92 90 94 96 97 90 00 101 102 03 Th Pa U Np P Am Cm Bk f Es m Md N Lr 2320 2310 2300 237 244 243 247 247 251 252 257 250 259 262 8314 JmolK Ical 4184 lLatm 101325 Water data 008206Latmm01K AHfus 334 Jg 1987 calmolK 1m 760m AHmp 2260 Jg 0 C 27315K ice 209 Jg C f X 34 139 254 d d 3 spam 1c 66262 10 15 1n cm e ne water 4184 Jgu C J heats Cs 300 x108 ms 11b 4536g steam 203 Jg C 96485 Cmole 1gal 3785L LECTURE 6 THE FINE POINTS IN PERIODICITY FILLED AND HALF FILLED SHELL STABILITY Summary If ENC was all we needed to explain trends in the periodic table then life would be easy But we all know that life isn t easy And in fact if we start to scratch to much at the surface of things we uncover an underbelly that we probably wish we hadn t seen So if we let the surface be the nice smooth trends of ENC and we let the underbelly be lled and half lled shell stability we are staring at the real data for electronic con gurations and for periodic trends You see rules like Auibau and ENC are really pretty and give us very straight lines But the reality is that most of the data has a bunch of bumps in it that create secondary effects we need to explain f mMA RCMP a 5amp7 M a r11th am 5 a 23 sq m T5 M M 27uwultm So what are some of these secondary bumps that need to be explained electronic con gurations of atom don t always follow Aubau example Cu is sld10 not s2d9 electronic con gurations of ions don t always follow Aubau example Tl 3 is d10 not s2d8 ionization energy trends are not smooth across a row of the periodic table example IE for O is less than IE for N electron af nity trends not smooth across a row of the periodic table example IE for C is greater than IE for N The good news is that once we understand the concept behind lled and half lled shell stability we can rationalize away the examples above and others like them with con dence Stable lled and halH Illed shells We all know the noble gas elements He Ne Ar Kr Xe are stable They are unreactive They are low energy orbitals They don t like to change What they L 39 39 lled s orp b H e 4g s Ivarqu x 5 wt r tz Y1 What is interesting is that there are other islands of stability in the periodic table Groups of elements for which it is also true that mine 39 quot 39 i 39milai quot39 And they all share in common being lled or half lled I ll ll ill4i S E9 at 37 all SmLLLellS 7 HHL Cl it U1 r 99 a3 palmt use 4r 4 P Hint During your examination ofthe periodic table you should start to see certain groups or columns as just being special In ame way you will learn to look at the last column ofthe periodic table the szp6 group we call the noble gases as special you will learn to look at s2 and p3 and p6 and d5 and d0 as special too Not as special as p6 but more special than the other columns It is these special or stable columns in which the glitches in the ENC argument Will be found So it you are Working a problem and nd yourself in the Vicinity of on e of these groups pay attention Shell stability messes with Electronic Configuration Example 1 the d4 and dg cases Certainly the most famous example of increased shell stability messing up an argument is are usin the Aufbau principle to ll an electronic con guration For examp e electronic con gurations that change W en e because of lledhalf lled shells are all the d4 and d9 electron con gurations In each case a single electron in the s2 orbital is relocated to the d5 or d orbit because of the increased stability in lling or half lling a large d subshell 9 MM SW Mtt 4 5 x a 39 Mg t 4 D Examples of this exception Which applies to not only atoms but also ions With the same number of electrons 0 Cr Mo W and ions like Fell Ruzl Ir all provide an electron from s2 to create a sld5 electronic con guration 0 Cu Ag Au and ions like lel An all provide an electron from s2 to create a sldm electronic con guration Example 2 the large metals with valence p1 to p4 electrons You may not have noticed that there are a collection of metals at the bottom right of the periodic table They aren t the famous transition metals like iron or copper But they are pretty important and actually possess some pretty remarkable properties This includes some of the so er metals like tin indium and lead among others The cations of these soft metals need to lose electrons and the question is from what subshell are they lost Accroding to Aufbau they should come from the valence p orbits But actually the subshells for these larger ions are so far from the nucleus that the simple Auibau energy rules fall apart And instead the question is Should they come from an s or a p or a d subshell The answer Remove them so that a filled d10 can be created as soon as possible So the rule is electrons lost from large ions come off in the following order p first s second d last Example anr loses its electron from p orbit e configuration to make a dlos2 In3 loses its electrons from p first and then the s orbits to make a d10 The pictures below show off the electronic configurations for these exceptions Remember all this because we love dw I 55 2 W Kr4d 552 3 In3 Kr4d Shell stability messes with Periodic Trends Recall the phrase to the right and up It suggested that we should see beautifully smooth trends that followed ENC 112 The the reality is that except for atomic radii this kind of smoothness in the trends just doesn t happen And instead we see a lot of jaggedness in the pictures of ionization energy or electron af nity Can we eXplain this jaggedness Well since we are in the section on shell stability this had better be the reason miveammic mam a MomicnumbevJ So here is something exciting to do Try staring really closely at the wiggles in the IE and EA data n lt 4 o r u m m m vmuoo mamwam manh sum w m m m mu moans um trauma9 m soc m m w m memo IISIIHSquY x ma 4 m m m no looIna mm lumr What you should notice is that the seemingly random up and down actually occurs around the islands of stabilityithe lld and half lled shells and subshells In the blow up below of IE data notice that the group to the right of a stable group is higher in energy less stable 139 kn 109 Ranking IE or C N and O The famous examples of this phenomenon o en found on exams occurs for the stable half filled p3 subshell Consider what is going on with electron configuration aroun p2 3 an p4 and ask how would you rank IE for C N and O The answer is CltOltN NOT CltNltO Why Look at which configurations are most stable 1 C is lower than N because N has higher ENC N is higher than 0 because p3 doesn t want to lose electron 4 4L 4 F3 O is lower than N because p4 likes to become p3 offsetting ENC 4 439quot VS 4 ENC F l Ranking EA for C N and 0 Another famous examples of this phenomenon also found on exams occurs for the stable half filled p3 subshell Consider what is going on with electron configuration aroun 2 3 and p and ask how would you rank EA for C N and O The answer is NltCltO NOT CltNltO Why Look at which configurations that are most stable 2 4k 7 F C has a more positive EA because adding e makes p3 F 3 39IL 9 9 F N has a more negative EA because adding e makes p4 g I I FW 0 has the most positive EA due to higher ENC And if you want to see the numerical proof of these examples if is presented below C N O F IE 1090 1400 1310 1680 EA 122 7 141 328 The hiccup in the smooth trend suggested by ENC is directly attributed to the islands of stability around p339 Ranking the islands of stability No doubt you will wonder some times just whether to look for exceptions and the magnitude of the stability increase is going to matter Is there a way to quantify how much extra stability is realized Sure program a computer with the Schrodinger equation But barring that can we come up with a ball park idea of how important the added stability is Sure ObViously the nobel gas configuration is way more stable than anything else And based on the examples we have seen p6 gtgtgtgtgtgtgtgt d10 gtgtgtgt d5 gt p3 gtgt 52 Relative extent of stability from filled and half filled shells llO 111 CH301 Fall 2008 Worksheet 2 Answer Key 1 From memory to the best of your ability write all regions of the electromagnetic spectrum discussed in class in order from longest to shortest wavelength TV radio microwave infrared visible ultraviolet Xray yray cosmic rays 2 Now write the same regions of the electromagnetic spectrum in order from highest to lowest energy Cosmic rays yray Xray ultraviolet visible infrared microwave radio TV Remember E hv hc7t In other words energy E is inversely proportional to wavelength 7 and so longer wavelengths correspond to lower energies thus the order is reversed relative to question 1 3 If asked to write the spectrum in order from highest to lowest frequency would they be written in the same order as in question lor question 2 Explain your reasoning The spectrum would be written in the same order as question 2 Remember E hv In other words energy E is directly proportional to frequency v and so high energies correspond to high frequencies thus the order would be the same as in question 2 4 Consider light with a wavelength of 470 nm a What color of visible light would this correspond to Blue b What would the frequency of this light be 470 nm 4 70x10397 m v cx 3x108 ms391470x10397 m 638x1014 s391 638x1014 Hz c How much energy would a single photon of this light posses Express your answer in joules E hv 6626x103934 Js638x1014 s391 423x103919 J d How much energy would a mole of such photons posses Express your answer in kilojoules per mole Remember we just calculated a single photon s energy in joules so just multiply by Avogadro s number to get joules per mole and then convert to kilojoules per mole You can think of it like this 423x103919 Jphoton3916022x1023 photonsmol391 254692 Jmol391 250 ld39mol39l 5 If the average bond energy for the 0H bond in water is 4589 kJmol391 what is the minimum frequency of light necessary to break that bond This is just the reverse of the two calculations done in question 4 parts d and c 4589 250 kJmol3916022x1023 photonsmol391 7620x103919 J39photon39l v Eh 7620x103919 J6626x103934 Js 1150x1015 s391 1150x1015 Hz 6 What failed prediction made by classical mechanics about the nature of electromagnetic radiation lead to the ultraviolet catastrophe Classical mechanics predicted that the power radiated by a blackbody radiator a fancy name for a hot object that radiates all frequencies of electromagnetic radiation would be proportional to the square of the frequency at which it emitted radiation and thus approach in nity as the frequency increased This was obviously not the case 7 at higher frequencies like ultraviolet blackbody radiators emit less not more let alone infinite power This was a catastrophic failure of classical 39 39 Whence quot 39 39 39 r 7 What is the photoelectric effect How did it demonstrate the inadequacy of the classical mechanical description of light The photoelectric effect was the observation that when light was shined on a metal surface electrons were ejected from the metal Classical mechanics predicted that the energy of the ejected electrons would be proportional to the intensity of the light regardless of its frequency Experimentally however it was observed that a minimum frequency energy of light was required to eject a photon and that the energy of the ejected photon was proportional to the energy of the incident light but not the lights intensity Classical mechanics fail 8 What is the de Broglie wavelength of a standard round fired from a Mauser C96 A standard round would have a mass of 56 g and an initial velocity of 430 ms391 56 g 56x10393 kg x hp hmv 6626x103934 Js56x10393 kg430 ms391 28x103934 m 9 What would the relativistic hypothetical mass of a photon of red 700 nm light in a vacuum be 700 nm 700x10397 m m hOtv 6626x103934 Js700X10397 m30X108 ms391 316x103936 kg 10 Using the relativistic mass of the photon calculated above and the equation E mcz calculate the energy of that photon Compare that value to the value derived using Planck s relation E mcz 316x103936 kg30X108 ms3912 284X103919 J E hv hcx 6626x103934 Js30X108 ms391700X10397 m 284x103919 J The two values should be exactly the same if you didn39t round anywhere You39ll notice that since E is one value for a given photon you can state that mc2 hc7t which conveniently rearranges to the de Broglie equation 11 What is the minimum uncertainty in the position of an electron mass 911XlO3931 kg traveling at a velocity with an uncertainty of30XlO7 ms391 AXAp 2 h4TE AXmAv 2 h4TE AX 2 h47ImAv AX 2 6626x103934 Js47I911X103931 kg30x107 ms391 AX 2193x103912 m 12 If scientists wanted to calculate the mass of a newly discovered subatomic particle for which they had only position AX 3313X10399 m and velocity Av 20XlO6 ms391 data what would that mass be Why is this a bad method to calculate a particle39s mass m 2 h47IAXAv m 2 6626x103934 Js4n3313x10399 m20x106 ms391 m 2 796x103933 kg This is a bad method to calculate mass because the uncertainty principle is an equality and so this calculation only establishes a lower limit for the mass of the particle which could potentially be much more massive 13 From memory to the best of your ability list all possible values the boundary conditions for the quantum numbers n 1 m1 mg n can have any integer value from 1 to infinity Z1 00 I can have any integer value from 0 to 711 Z0 n 1 M can have any integer value from 1 to l Zl l my can have a value of either 12 or 12 14 Determine whether the following sets of quantum numbers are valid If you determine they are invalid explain your reasoning an3l0m10ms12 Valid bn3l2m12mx1 Invalid 1 is not a possible value for my cn0l0m10ms12 Invalid 71 cannot be zero it can only have integer values from 1 to infinity dn2l2m11mx12 Invalid I cannot be equal to n it can only be less than n en4l3m14m512 Invalid M has integer values from Z to 1 4 is greater than 3 and thus invalid 15 Suggest possible values for quantum numbers that could fill in the blanks below without violating any boundary conditions an3l0m1imx12 0 Becauselis 0 m can only be 0 bn6lim13mx12 34or5 Because 71 6 could be 0 1 2 3 4 or 5 but because m1 3 must be at least 3 cn2l 1 m11 mxi 12 or 12 Because my can only have a value of either 12 or 12 it doesn39t matter what the other quantum numbers are 16 State the Aufbau principle in your own words Without consulting a table your notes or any peers write the Aufbau order Hint you can read the Aufbau order right off the periodic table The Aufbau principle states that electrons will ll lower energy orbitals before lling higher energy orbitals ls 2s 2p 3s 3p 4s 3d 4p etc Remember you can read this right offthe periodic table Just apply the fact that p orbitals start at 2p d orbitals at 3d and f orbitals at 4f 17 State Hund39s rule in your own words In the simplest terms Hund s rule states that all degenerate orbitals in a given subshell should be halffilled with electrons before any degenerate orbital is completely filled with electrons 18 State the Pauli exclusion principle in your own words Any valid set of 4 quantum numbers can describe one and only one electron in a given atom or ion likewise each electron is described by one and only unique set of 4 quantum numbers 19 Using quantum numbers describe a The highest energy electron in a Boron atom n2l 1 m1 1 mx12 b The lowest energy electron in a Xenon atom 71 110m10mx12 20 What is an element that can have a ground state electron described by the quantum numbers n7l 1 m1 1 mx12 Ununtritium Uut or any element beyond it in the periodic table is an acceptable answer


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