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Exam 3 Study Guide

by: Brittany Notetaker

Exam 3 Study Guide CHE140

Brittany Notetaker
GPA 4.0
General Chemistry I
Dr. Chris Hamaker

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Here's a study guide to help you on the exam this week! Good luck!
General Chemistry I
Dr. Chris Hamaker
Study Guide
CHE140, Chemistry
50 ?




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This 25 page Study Guide was uploaded by Brittany Notetaker on Sunday October 25, 2015. The Study Guide belongs to CHE140 at Illinois State University taught by Dr. Chris Hamaker in Summer 2015. Since its upload, it has received 64 views. For similar materials see General Chemistry I in Chemistry at Illinois State University.


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Date Created: 10/25/15
CHE140 Exam 3 Study Guide Chapter 7 the distance from crest to crest from maximum to maximum i Wavelength L ed 5 5t V reugh r I the number of waves per unit time usually seconds Higher Frequency Eherter Wavelength Lewer Frequency enge Weeellngth Highest gt Energy gt Lowest Highest gt Frequency gt Lowest Shortest gt Wavelength gt Longest Atomic Spectra 0 Emission AKA bright line spectra 0 Electrons in atoms are excited and give off their extra energy as light electromagnetic radiation e gtlt excited 6 Add energy Light is given off at specific wavelengthsenergies C e 0 Absorption AKA dark line spectra 0 When continuous light is passed through atoms of an element only certain wavelengths are absorbed I The emission and absorption wavelengths are the same for a given element I Different elements have different atomic spectra Continuous Spectrum Emission Spectrum Absorption Spectrum Quantum Theory 0 the smallest amounts of energy emittedabsorbed o a light particle the energy of one quantum of light 0 Planck proposed that the energy E of a photon was directly proportional to the frequency of the wave of its corresponding radiation E hv 0 v is the speed of an object c is the speed of light 0 v and c are interchangeable o The energy of a photon is inversely proportional to its wavelength E hck DeBroglie Wavelength 7 hcE E mc2 Substitute mc2 for E k hcmc2 The c s cancel out 9 hmc Substitute c for V k hmv 1 What is the deBroglie wavelength of a 95mph fastball 95mph 4247 ms Mass 145g 0145kg h 6626 X 10 34 JS 1 1kgm2s2 substitute kgm2s2 for J one of the 5 s will cancel out h 6626 X 103934 kgm2s k hmv 6626 x 103934 01454247 I Highlighted items cancel out leaving m on top to equal 108 X 103934 m 108 X 103925 nm Visible light is in nm Heisenberg Uncertainty Principle 0 We cannot know exactly the speed and position of an electron o For example we know the speed of a car as it is driving but not the exact location 4 Lines in Visible Region Hydrogen o Johan Balmer v 32881 x 103915sl22 ln2 o ngt2 o Paschen IR lines V 132 1n2 o ngt3 o Lyman UV lines V 112 1n2 o ngtl o n energy levels of the electrons in hydrogen 0 Rydberg was able to revise these equations An atom in its lowest possible energy level is in its If an atom is in an energy level above it is said to be in an When an electron in an excited state moves to a lower energy state or ground state it emits a quantity of energy that exactly matches the energy differences between two states This change in electron energy is called an Quantum Numbers and Electron Spin 0 Schrodinger Equation 0 Each electron has its own wave function w which de nes energy levels 0 1 2 is a probability function 0 It de nes an orbital an area where you re most likely to nd an electron 0 Has 3 unique integer values called quantum numbers Principal Quantum Number 11 0 Same as Bohr s n 0 Qrbitals of the same nvalue are in the same shell 0 11 defines a shell 0 The larger the n the farther away from the nucleus meaning a larger orbital Angular Momentum Quantum Number 1 o Integer value from 0 nl o Defines a subshell n and l 1 Letter subshell Magnetic Quantum Number m1 0 Defined by l with integer values of l to l 0 Describes its orientation in space of the orbital which direction does it point 0 The number of orbitals in a shell is equal to 21 1 o The number of orbitals is equal to n2 n 1 m1 Notation Orbitals 1 0 0 ls 1 2 0 0 2s 4 1 101 2p 4 3 0 0 3s 9 1 101 3p 9 2 21012 3d 9 4 0 0 4s 16 1 101 4p 16 2 21012 4d 16 3 3210123 4f 16 4th Quantum Number ms 0 Spin magnetic quantum number mS 12 0 No two electrons in an atom can have the same two sets of quantum numbers 0 n 1 m1 9 describe an orbital o 2 electrons per orbital o n 1 m1 ms 9 describe a specific electron Shapes and Sizes of Orbitals See httpWinter groupshefacukorbitron Electron Configuration 0 Placing electrons into orbitals o 2 electrons per orbital 0 Fill from lowest energy to highest energy Hydrogen 1e 9 H Helium 2e 9 He 1s2 Boron 5e 9 B 1s22s22p1 Neon 10e 9 Ne 1s22s22p6 Electron fill order 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p Sulfur 16e S 1s22s22p6 0 Core electrons Abbreviated Noble Gas Core Electron Configurations Vanadium 23 V 1s22s22p63s23p64s23d2 o Argon s con guration V Ar4s23d2 V Ar3d24s2 Valence Electrons 0 The outermost largest n electrons 0 For elements in groups 1 2 1318 only main group elements Ge 1s22s22p63s23p64s23d104p2 Ge Ar4s23d104p2 Ge Ar13d1039 Be He2s2 Mg Ne3s2 Ca Ar4s2 0 Groups have the same number of valence electrons Orbital Diagrams 0 Fill orbitals using Hund s Rule 0 Maximize spin 0 Place an electron in each degenerate before pairing T1 T1 TllTllTl Tl TllTllTl TllTllTilTilTl 15 25 3s Tl TllTlltl TllTllTlITlITl Tl MW 45 5S TllTllTllTllTllTllTl TllTllTllTllTl Tl TllTllTl 65 o Exceptions o Chromium 9 Cr Ar4s13d5 Clquot I 0 Copper 9 Cu Ar4s13d10 I lL lL lL lL II 4s 33921 m Sr Kr5s2 Sr2 Kr 0 Lost two valence electrons Rb Kr5s1 Rbz Kr 0 Lost one valence electron o the same electron configuration Br Ar3d104s24p5 Br39 Ar3d104s24p6 Kr 0 Br gained one electron therefore it has the same electron configuration as Kr Transition metals lose their s electrons first to form cations Mn Ar4s23d5 Mn2z Ar3d5 Periodic Trends Atomic Radius Covalent Radius 9 The atomic radius is the size of an atom VI 2 1 me i V 394 5 39L39 u 39939 to J Be B C F Ne I I 12 39 l I I i Iquot it 1 8 h a Mg Al Si P Ci Ar V I 10 3 ll 2 121 2T Ell Ill 3239 Ill 3 5 t 2639 K Ca Sc Ti Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr i 3 1 JOI ai 39 I 43 I u39 is 1 u 43 419 so 39 1 32 395 J Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd l n Sn Sb Te I Xe I 3 5m 7i 72 i i u 7 77 i9 3 3982 ii will ilf it Cs Ba Lu Hf Ta W Re 05 lr Pt Au Hg Tl Pb Bi Po At Rn I a 31339 9402 ml nu in i39 i in 1st un 39u39o 39m39 Hi2 39 til I A I Fr Ra 4 Lr R f Db Sg Bh Hs Mt Uun Uuu Uub Uuq uo til ca 55 a quot Ce Pr Nd Pbm Sm Eu Gd Tb Dy Ho Er rm Yb 3 333 3g s9 V 9139 3992 A91 V i 5 9539 9quot 6 int riozt 5 ii Pa U NP PF 5quotquot 5 393 9f 5 Fm 39Y39 N o The atomic radius increases from top to bottom 0 Atoms get larger as we go down because you are adding shells 1s 2s 3s etc 0 The atomic radius increases from right to left 0 Atoms get larger as you move right to left because you add electrons from left to right and the charge on those electrons lessen H He 1 lt5 53 31 Metals B C N O F Ne 2 2 35 P n Semimetals U 4quot J39 quotVquot 167 1 1 2 m 87 67 56 48 42 38 1 Nonmetals 3 O o m 190 145 118 111 98 88 79 71 xi 5 mi 4 0 G 243 194 184 176 171 166 161 156 152 149 145 142 136 125 114 103 94 88 U I 265 219 212 206 198 190 183 178 173 169 165 155 156 145 123 115 108 133 6 La OOO 298 253 208 200 193 188 185 180 177 174 171 156 154 143 135 127 120 o Electrons in lower levels can screen the outer electrons o Shielding the proton does not feel the last electron in the outer shell 0 2eff 9 the positive charge an electron feels this is the effective nuclear charge Ionization Energy 0 Ionization energy is the energy it takes to remove electrons go off of the first ionization energy V l quotI I 1 s s quotiquot a o to ll Be 8 C N O F Ne quot1quot quotTiquot 1quot quotquot62quot quot1 1397quot 391 quot Na Mg Al SI P S Cl Ar quotifquot 3951 quot395quot i 3923quot quot 23quot 3395 33quot quot3392quotquot 5 quotE39squot K Ca Sc TI V Cr Mn Fe Co Ni cu Zn Ga Ge As Se Br Kr 1 quot5399quot quot2393quot 1339 quot3quot 39139quot 0 quotquot39 5 5quot Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te 1 Xe In in u 1 J u u u 1 m n w n w 39 H II III 4 v A hr 3923 thy 1I l39v 5710 7 quot quotquot13quot quot 77quot quot397 quot 3quot quot7 7quot 13 quot15quot quotaquot h u 39u 5 u Cs Ba Lu Hf Ta W Re 05 Ir Pt Au Hg Tl Pb BI Po At Rn quot 39 39 3939 39 2 t quot1 139quot h 9402 1331 I06 106 101 I m quot0 111 quot2 quot4 Fr Ra Lr Rf Db Sg Bh Hs Mt Uun Uuu Uub Uuq 39 391 mi L39 39l 139 Ir wl L39 quot1 l739 39l m J mg 1 39l 1 n 5quot 39uEo nn reggun 11631 lsginf qua7quot 4amp2 66 6 39 JGgquot 1373 L Ce PT Nd Pr SFquot E 99 TP DY 11 FF T Y 39 gr 56quot 39 H 39i 53 V islquot quot9395quot Iquot aquot mg 100 0 1052 Ac rh Pa u Np Pu Am cm Bk Cf Es Fm Md No L 39Lanthamde senes lr39 mm 1nv u u I 1 4 v a n v n n 0 Elements on the bottom are the easiest to remove electrons from and get more difficult as the elements move upward 0 Elements on the left are the easiest to remove electrons from and get more difficult as the elements move right On13ho 639 0 11m w 2500 0 Period 2 has some anomalies Element Li IE1 520 1086 F Ne J 1681 2081 0 Generally the ionization energy increases but there are two anomalies in period 2 o The change between Be and B is negative as well as the change in N to O o Be s abbreviated electron con guration is He2s2 whereas B s is He2sz2p1 B requires less ionization energy because it is easier to pull electrons from a shell that is not full 2p1 compared to a shell that IS full 2S2 0 Beyond IE1 IE increases Traditional IE1 IE2 IE3 IE4 Charge 1 Na 496 4560 6910 9540 2 Mg 738 1445 7730 10600 3 A1 578 1815 2740 11600 o Removing electrons past their traditional charge requires more energy trying to remove from a full shell Electron Affinity quot5 9 5 o 7 9 lo B C N F Ne if quotifquot if Wfo39quot quotT7quot vr i39a39 639 3 P S 039 AT 25 3922 quot2139 quot 5 I quotaquot 39 quotquot 3239 quot5quot quotsquot 55quot K Ca Sc Ti Mn Fe Co Ni Cu 2n Ga Ge A39s Se Br Kr 7 3quot 7amp4quot Ha 39J if quotquot735quot 23quot quotif isquot quot 15quot 5039 quotaquot 52quot 2139 E Rb Sr Y Zr Nb Mo 139 Ru Rh Pd Ag Vlln Sb Te I Xe 539 mo 397 39quotnquot 3 quotfigquot 7773 7quot23977397 is 5quot 3956quot quot 1quot quot839 3933quot 3911quot quot3935quot E39squot 25 Ba Lu Hf Ta Re Os Ir Au Hg Tl Pb Po At Rn 839 quot402 I39DJ yIlltly7r is lllh Iii v1quot A I h V A 3r Ra hr R f Db Sg Bh Hs Mt uunUuuU ub Uuq m 585 s r sis quotisquot quotco quot395quot z39 539 o s39 quot39m39 2 o a o 6 VLa Ce Pr Nd Pm Sm Eu Gd Tb Py Ho 3 Tm Yb VLclnz Sfi3 39i a 9 9 29 394 9396 l IIEHW 3993quot 5 161 391 0139 JlOZ H AcTh PaUNpPuAmcmBk cfEsFquNo 0 Going up a group the atoms typically get more negative 0 Going from left to right in a period typically the atoms get more negative n or F 33 141 4523 3 11 J2 m i 341 4393 495 25 31 TE 1 A 133 491 E 5 55a 63 Em Rules 1 Group 18 all have positive endothermic electronic affinity 2 Halogens all have very large negative values they REALLY want an extra electron 3 Group 1 to Group 2 is the biggest difference Atomic Size vs Ion Size 0 Cations are smaller than the parent ion 0 Anions are larger than the parent ion 0 Why Z protons stay the same but the number of electrons change Na 1 lp 1 le Na 11p lOe makes a cation loses and electron size decreases S 16p 16e Sz39 16p l8e makes an anion gains two electrons size increases Isoelectronic Ions 0 Same electronic configuration 0 The proton electron ratio determines how large or small the ion will be 0 The larger the proton electron ratio the smaller the ion will be 0 The smaller the proton electron ratio the larger the ion will be 02 F Na Mg2 p protons 8 9 11 12 e electrons 10 10 10 10 Size pm 140 133 102 72 0 Mg2 is the smallest ion because the number of protons compared to electrons is the greatest 0 0239 is the largest ion because the number of protons comparted to electrons is the least Types of Bonds Ionic Bonding bond between a metal and a nonmetal o Electrostatic attraction between oppositely charged ions 0 One atom gives up an electron to the other Metallic Bonding bond between metals exclusively o The atoms electrons are pooled together creating bands causing conduction electrons ow through the bands Covalent Bonding 0 Covalent bonding is the sharing of electrons 0 As the atoms move close to each other the nucleus of atom 1 feels the electrons of atom 2 and vice versa 0 Eventually the two nuclei will repel each other 43E Fetentiel energy Mime 4 G G I e AIDE Energy rel eeee lenne 7 Fund Energy when bend 4132 45Equot 0 At the minimum the attractive electrostatic force is maximized H H2 band Eng Internu clearquot diele nee nrnjl 0 Distance at the nucleus is the bond length 0 Energy at the minimum is the bond strength 0 Energy is given off when you make bonds 0 Lowering the energy of the electrons in covalent bonds Lewis Symbols 0 Represent the valence electrons in an atom One dot per valence electron the number of preferred bonds an atom will make is equal to the number of single dots Paired electrons Unpalred Unpaired electrons electron 39 39Cl C o o o O O O O Chlorine Carbon Callyughl 39 Pearson Fl lucanon mo ullshmr1 as Pryarson AUUIBDH V JHS KPV o The goal of covalent bonding is to reach an octet 8 valence electrons 0 Lewis Structures are 2D representations of molecules showing how atoms are bonded together and how the electrons are distributed 0 Electrons are in pairs 0 are electrons shared in a covalent bond 0 belong to only one atom in the molecule 1 NF3 a First determine the total number of valence electrons available i N 1 X 5e 5e ii F 3 X 7e 21e iii Total of 26e or 13 pairs b Second arrange the atoms as they are bonded together and add single bonds i This is represented by dashlike lines 2e per line c Next complete the octets of the outer atoms i Outer atoms are F d Lastly add the remaining electrons to the central atoms as lone pairs i Central atom is N e If the central atoms does not have an octet move outer lone pairs to make multiple bonds can be double or triple What about Polvatomic Ions 0 Each positive charge is one less electron to the total 0 Each negative charge is one more electron to total COC12 i39D39I II II ii i f H H C032 2 C03 2 2 2 O C quot c o 0 H Q l H 0 0 O 9 O 1 o no u o 0 C03239 has resonance the true structure is an average of all possible resonance structures 0 The average is a 1 13 CO bond Electronegativitv o Electronegativity is the ability of an atom to attract electrons to itself in a covalent bond ke 339 a gear Nquot n51 Xi i fr 2 a g 2 a 5 t 5 239 a a A 9 a 2 21 9f 392 at 11 2 21 5 52 Kg 3 r 398 quot f2 2 5 Bi Li a 2 or a I 339 4 3 439 5 a at 25 s aquot 3 fr 3 a 3 s 3 it as as as Ea C Er Nd Pbm Sfquot Eb Gd Tb Dy H6 Tm Yb Ac Th Pa U Np Pu Am Cin B k 5f Es Fm Md quot2 N9 o The electronegativity trend electronegativity increases as you move up the table and to the right of the table 0 The most electronegative elements 0 F 40 o O 35 o N C1 30 o Electrons in covalent bonds are not always equally shared 0 Bond polarity 0 Polar covalent bonds 0 Unequally shared electrons 0 Different electronegativities o NonPolar covalent bonds 0 Equally shared electrons O AEN 0 C S 0 Both have the EN of 25 0 AEN 0 0 Non polar bond F F 0 AEN0 C 0 0 C has an EN of 25 0 Less electronegative meaning it has the electron less often giving it a slight positive charge 0 O has an EN of35 o More electronegative meaning it has the electron more often giving it a slight negative charge 0 AEN 10 0 Polar bond 0 The larger the difference in electronegatiVity the more polar the bond 0 The smaller the difference in electronegatiVity the less polar the bond Exceptions to the Octet Rule 1 An odd number of e overall 0 The atom With the lowest electronegatiVity gets the odd electron and is therefore short of an octet one 2 Boron usually has less than an octet i i i 9 DH DIR 9 DH 9 H B H o BH3 has 6 valence electrons 3 An expanded octet has more than 8 electrons 0 Only elements With equal to or more than the atomic number of 12 0 Elements 12 P C I Phosphorus Pentachloride O o o o C O azPV o O O o P has 10 valence electrons 0 Expanded octets usually only expand to 10 or 12 valence electrons X e Xeno39 tr uoride gt 70 00 039 o FeF on 0 g 00 E o Xe has 12 valence electrons Bond Length and Bond Strength Bond Length pm BDE kJrnol C O 143 358 C O 123 743 C O 113 1072 o Bond Dissociation Energy H Longest lengtl weakest energy 1 H coH l Estimating AH using BDE s Aern ZAHbreaking 39 ZAHforming C2H4 302 9 2CO2 2H2O Breaking H H X gx CC Making o cio Shortest lengtl strongest energy Broke 4 C H bonds 1 CC bond 3 00 bonds Making 4 C20 bonds 4 O H bonds I I HOH II AHrxn 4C H 1CC 300 4CO 4O H 4413 kJmol 614 kJmol 3495 kJmol 4799 kJmol 4463 kJmol Aern 1297 kJmol exothermic Shapes of Molecules C02 I I I I39 9 C 9 0 C02 is linear 0 C SN 2 S02 33 Ej 0 SOzisbent o SSN3 Valence Shell Electron Pair Repulsion Determines the shape Electron Pair Geometry Parent Geometry o EPG is the positions of lone pairs and atoms 0 Determined by the steric number SN Molecular Geometry Shape o Determined by the positions of atoms Steric Number SN SN of lone pairs of atoms bonded A central atom X bonded atoms E lone pairs SN 2 EPG Linear 0 MG linear I I I I 9 C 9 SN 3 EPG Trigonal Planar 0 3 atoms 0 lone pairs AX3 0 MG trigonal planar IH H H 1200 angle between bonds 0 2 atoms 1 lone pair AXzE 0 MG bent H j j39 I A little less than 1200 between bonds Lone pairs occupy more space than bonding pairs Multiple bonds occupy more space than single bonds Lone pair gt Double bond gt Single bond SN 4 EPG Tetrahedral 0 3 atoms 1 lone pair AX3E 0 MG Trigonal Pyramidal o o N lquot I I Bond angles are approx 10950 0 2 atoms 2 lone pairs AX2E2 0 MG Bent G O G C 00 39039 0 pl SN 4 EPG Trigonal Bipyramidal 0 5 atoms 0 lone pairs AXS 0 MG Trigonal Bipyramidal q 0 4 atoms 1 lone pair AX4E 0 MG See saw II F Ii i Ii Ii i is i F BF F i I l i 39l i I F 1 l I 0 3 atoms 2 lone pairs AX3E2 0 MG T shaped F 435 F 11 b I I i G I 0 2 atoms 3 lone pairs AX2E3 0 MG Linear oo 00 o l l l O SN 6 EPG Octahedral 0 6 atoms 0 lone pairs AX6 0 MG Octahedral f quotquotI a Bond angles are 900 0 5 atoms 1 lone pair AXsE 0 MG Square Pyramidal F IFS fI Iodine Penta uoride o o 0 4 atoms 2 lone pairs AX4E2 0 MG Square Planar l 14 u I o F Polar Bonds and Molecules 0 Polar bonds have a bond dipole 0 One slightly positive end one slightly negative end 0 Polar molecules have a molecular dipole 9 one end positive one end negative 0 Molecules whose ends have the same charge are not polar Polar Molecules Need 1 At least one polar bond 2 Asymmetry a A lone pair on the central atom I More than one type of atom on the central atom S f H f Polar and Non Polar Molecules Atom Electron tivi C 25 H 21 Cl 30 N 30 l N Cl I 00 21 AEN O The electronegativity between N and Cl is the same Non polar bonds Non polar molecules H I39T39I H H AEN Polar bonds Polar molecules 39i il ij il C I C HAEN04 C CIAEN05 Vibratin Bonds and the Green House Effect Earth39s Greenhouse Effect Greenhouse gases let the sun s short wave radiation visible light reach the earth but trap some of the long wave infrared or heat radiation coming from the warm earth Incoming solar radiation Reflected by the short wave 9mm and its Infrared radiation atmosphere heat emitted from earth39s surface long wave Atmosphere Carbon Store mmhare E B l Solar radiation strikes the surface and heats it up through absorption With the surface heats up it emits IR radiation 3 Molecules can absorb IR radiation which causes bonds to Vibrate and increase their kinetic energy therefore they give off heat Warms the earth39s surface 591 0 Without the greenhouse effect Earth s surface would be about 18 C 0 The actual surface temperature is about 15 C o The main gases that absorb the IR radiation and heat up the surface temp are H20 CH4 C02 and CFC s o In order to absorb IR radiation 0 The compound must have polar bonds 0 Vibrations must change the molecular dipole Valence Bond Theory 0 Bonds are formed from the overlap of valence atomic orbitals 0 Sigma bonds the orbitals overlap between the nuclei SS 0 Pi bonds the orbital overlap does not occur between the nuclei but above and below the nuclei PP 0 of atomic orbitals 0 Atomic orbitals miX and average Tetrahedral EPG 9 sp3 hybridization 10950 angles ll all hybridization T T Spa energy 1 SP3 hybrid orbitals have one s orbital and 3 p orbitals Trigonal Planar EPG 9 sp2 hybridization 1200 angles C uc n v 1 1 J at p l c pivr c y 39 Electronic Cm gtnation of C 1T H T T l In gr Quad state Electronic co gLH39athll of C LT T 7 l A In an cuted state i 5 1 hybrldIzatnjr I If C ll l l l l 1 I Three 3955 unhybx Mixed t r39brldlzed Orbit3 orbatals SP2 hybrid orbitals have three hybridized sp2 orbitals and one unhybridized p orbital Linear EPG 9 sp hybrid 1800 angles it 1 i t t t hybridization T Spa energy 1 0 SP orbitals have two sp hybridized orbitals and two unhybridized p orbitals Hybrid orbitals make sigma bonds Unhybridized orbitals make pi bonds


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